Ch 13 & 14

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Calculate the molality of each of the following solutions. 4.90 g NaCl dissolved in 0.350 L of water

0.236mol/kg (My answer of 0.240 was accepted by Mastering)

At a certain temperature, the solubility of N2 gas in water at 2.04 atm is 48.0 mg of N2 gas/100 g water . Calculate the solubility of N2 gas in water, at the same temperature, if the partial pressure of N2 gas over the solution is increased from 2.04 atm to 6.00 atm .

141 mg of N2 gas/100 g water Henry's Law: S=k×P where P is the partial pressure of the gas, S is the solubility of the gas, and k is the Henry's law constant. S1=k1×P1 and S2=k2×P2 where k1 and k2 are the Henry's law constant at the partial pressures P1 and P2, respectively. Because K1 & K2 are the same constant (and therefore are equal) the following equation can be used: S2 = (P2/P1)(S1)

Calculate the activation energy (Ea) in kJ/mol for a reaction if the rate constant (k) is 2.7 × 10^−4 (M−1 s−1) at 600 K and 3.5 × 10^−3 (M−1 s−1) at 650 K.

166 kJ/mol The equation ln(k1/k2) = (Ea/R)(1/T2 − 1/T1) provides a convenient way to calculate the activation energy given a rate constant, k1, at temperature T1 and the rate constant, k2, at some other temperature T2.

Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0.848 g of biphenyl (C12H10) in 25.0 g of cyclohexane?

2.10 C The temperature at which a solution freezes and boils depends on the freezing and boiling points of the pure solvent as well as on the molal concentration of particles (molecules and ions) in the solution. For nonvolatile solutes, the boiling point of the solution is higher than that of the pure solvent and the freezing point is lower. The change in the boiling for a solution, ΔTb, can be calculated as ΔTb=Kb⋅m in which m is the molality of the solution and Kb is the molal boiling-point-elevation constant for the solvent. The freezing-point depression, ΔTf, can be calculated in a similar manner: ΔTf=Kf⋅m in which m is the molality of the solution and Kf is the molal freezing-point-depression constant for the solvent.

A certain second-order reaction (B→products) has a rate constant of 1.00×10^−3 M−1⋅s−1 at 27 ∘C and an initial half-life of 204 s . What is the concentration of the reactant B after one half-life?

2.45 M First calculate [B]0 (initial concentration), using the half-life formula for a second-order reaction. After one half-life, the concentration of B will be half of its initial concentration. For second-order reaction the equation is t1/2 = 1/(k[A]0). In this case it is 4.0902 M.

What is the rate constant of a first-order reaction that takes 242 seconds for the reactant concentration to drop to half of its initial value?

2.87×10^−3 s−1 (2.86*10^-3 s-1 was accepted)

A certain first-order reaction has a rate constant of 8.00×10^−3 s−1. How long will it take for the reactant concentration to drop to 1/8 of its initial value?

260. s Calculate the half-life for this reaction in seconds. (86.6 s) Given the equation Fraction remaining = (1/2)^n where n is the number of half-lives, determine how many half-lives it takes for the concentration to become one-eighth of its initial value.

Calculate the molality of each of the following solutions. 8.64 g benzene (C6H6) dissolved in 23.5 g carbon tetrachloride (CCl4)

4.70mol/kg (It should be 4.71, but 4.70 is what Mastering gives)

How many grams of solution must contain 1.7 g of NH4Cl to make a 35% NH4Cl solution by mass?

4.9g (You can either do 1.7/.35 or make a ratio of 1.7/x = 35/100 and solve for x.)

The activation energy for the reaction NO2(g)+CO(g)⟶NO(g)+CO2(g) is Ea = 125 kJ/mol and the change in enthalpy for the reaction is ΔH = -275 kJ/mol .What is the activation energy for the reverse reaction?

400 kJ/mol Draw the energy profile diagram, making sure to assign the values of Ea and ΔH. Finally, use this diagram to determine the energy of activation for a reverse-order reaction. The energy profile can be traced either in the forward direction or reverse direction when calculating the enthalpy of the reaction by taking care that the sign convention for the enthalpy change is followed. For the reaction NO2(g)+CO(g)⟶NO(g)+CO2(g), the energy of activation for the reverse reaction can be calculated as Ea(reverse) = Ea(forward) − ΔH.

A sulfuric acid solution containing 571.4 g of H2SO4 per liter of aqueous solution has a density of 1.329 g/cm3. Calculate the molarity of H2SO4 in this solution. Express your answer in moles per liter to four significant figures.

5.825M The molarity of a solution is the number of moles of solute divided by the solution volume in liters: molarity = (moles of solute / liters of solution) Calculate the amount of moles of H2SO4 using its molar mass as a conversion factor: moles of H2SO4 = 571.4g H2SO4 × (1 mol H2SO4 / 98.09g H2SO4) = 5.825 mol H2SO4 Substitute the quantities into the equation for molarity: molarity of H2SO4 = (5.825 mol / 1L) = 5.825M

Seawater contains 7.7×10^−3 g Sr2+ per kilogram of water. What is the concentration of Sr2+ measured in ppm?

7.7 ppm

What is the half-life of a first-order reaction with a rate constant of 8.40×10−4 s−1?

825s

Which of the following are correct for zero-order reactions? Check all that apply. The concentration of the reactants changes nonlinearly. A higher concentration of reactants will not increase the reaction rate. The rate of reaction does not equal the rate constant. A zero-order reaction slows down as the reaction proceeds. The units for the rate constant and the rate of reaction are the same.

A higher concentration of reactants will not increase the reaction rate. The units for the rate constant and the rate of reaction are the same. A zero-order reaction is independent of the concentration of the reactants. Zero-order reactions have the following features: The concentration versus time graph is a straight line with a negative slope; this negative slope represents the rate constant of a reaction. The rate of the reaction is equal to the rate constant. The units of k and the rate of reaction are mol⋅L−1⋅s−1. Photochemical reactions and surface reactions are examples of zero-order reactions.

What is the structural geometry of the NOF molecule?

Bent

B→2A The concentration of A increases, while that of B decreases over time. Therefore, B is a reactant and A is a product. Since the final concentration of A is approximately twice as high as the initial concentration of B, the mole ratio of A:B is 2:1. Thus, the reaction is B→2A.

A graph of ln[A]t vs time yields a straight but slightly horizontal line. Also a graph of [A]t vs. time of the same data yields a curved line. What can be inferred about the reaction and rate constant? The reaction is first order and the rate constant is large. The reaction is first order and the rate constant is small. The reaction is zero order and the rate constant is large. The reaction is zero order and the rate constant is small.

The reaction is first order and the rate constant is small.

Consider the data presented below. Time (s): ____ 0 _____ 40 _____ 80 ____ 120 ___ 160 Moles of A: _ 0.100 __ 0.067 __ 0.045 __ 0.030 __ 0.020 Determine whether the reaction is first order or second order.

This is the first order reaction. (Do ln[A]t and 1/[A]t and plot that data against time. Whichever results in a linear line determines whether it's a first or second-order respectively)

The half-life of a reaction, t1/2, is the time required for one-half of a reactant to be consumed. It is the time during which the amount of reactant or its concentration decreases to one-half of its initial value. Determine the half-life for the reaction in Part B using the integrated rate law, given that the initial concentration is 1.50 mol⋅L−1 and the rate constant is 0.0018 mol⋅L−1⋅s−1 . *Note: the reaction is zero-order*

t1/2 = 4.2×10^2 s The faster the reaction, the shorter the half-life for that reaction. The rate of the reaction is proportional to the rate constant; thus, the larger the rate constant, the shorter the half-life.

Consider the data presented below. Time (s): ____ 0 _____ 40 _____ 80 ____ 120 ___ 160 Moles of A: _ 0.100 __ 0.067 __ 0.045 __ 0.030 __ 0.020 What is the half-life?

t1/2 = 68.7 s (68.9 was accepted. It seems better to use the first and last data points than the first and second although neither got me 68.7)

At 1 atm, how many moles of CO2 are released by raising the temperature of 1 liter of water from 20∘C to 25∘C? S (mol/L) ____ P (atm) ___ k (mol/(L⋅atm) ___ T (∘C) 3.70×10^−2 __ 1.00 _____________________ 20.0 9.60×10^−2 ___ ___ _____________________20.0 ____________ 1.00 ______ 3.50×10^−2 _____25.0

0.0020 mol (@20.0C S = 0.0370 mol/L @25.0C S = 0.0350 mol/L Therefore 0.0020 mols are released)

For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected: Trial __ [A](M) __ [B](M) __ [C](M) ___ Initial rate (M/s) 1 _____ 0.10 ___ 0.10 ____ 0.10 _____ 3.0×10^−5 2 _____ 0.10 ___ 0.10 ____ 0.30 _____ 9.0×10^−5 3 _____ 0.20 ____0.10 ____ 0.10 _____ 1.2×10^−4 4 _____ 0.20 ____0.20 ____ 0.10 _____ 1.2×10^−4 What is the reaction order with respect to B?

0 The rate didn't change when [B] was doubled. This indicates an order of 0 with respect to B. Compare the values in trial 3 to those in trial 4. Notice that [A] and [C] remain constant whereas [B] doubles. What happens to the rate in the transition from trial 3 to trial 4? It stays the same. The reaction rate can be determined from the following equation: rate = k[A]^x[B]^y[C]^z where y is the order with respect to B. For trial 3 this equation is: 1.2 × 10^−4 = k(0.20)^x(0.10)^y(0.10)^z and for trial 4 this equation is: 1.2 × 10^−4 = k(0.20)^x(0.20)^y(0.10)^z Dividing the equation for trial 4 by that for trial 3 gives (1.2×10^−4) / (1.2×10^−4) = (k/k)(0.20 / 0.20)^x(0.20 / 0.10)^y(0.10 / 0.10)^z which simplifies to 1 = (1)(1)^x(2)^y(1)^z Now, solve for x (This can be done by dividing the log of the left side by the log of the right side. You have 1 = 2^y. Take log(1) and divide it by log(2) to get the value of the exponent.).

https://session.masteringchemistry.com/problemAsset/1061737/43/BLB-1061737-graph.jpg A solid mixture consists of 40.8 g of KNO3 (potassium nitrate) and 7.2 g of K2SO4 (potassium sulfate). The mixture is added to 130. g of water. If the solution described in the introduction is cooled to 0 ∘C what mass of K2SO4 will crystallize?

0 g (The solubility of K2SO4 at 0C is about 8g/100g of water. The K2SO4 is added to 130g of water. 8g of K2SO4/100g of water X 130 g of water gives 10.4 g of K2SO4 that is capable of dissolving at 0C. Because 10.4 g can be dissolved at 0C and only 7.2 g were dissolved that means all of the K2SO4 remains dissolved and 0 g crystalizes.)

What is the Henry's law constant for CO2 at 20∘C? S (mol/L) ____ P (atm) ___ k (mol/(L⋅atm) ___ T (∘C) 3.70×10^−2 __ 1.00 _____________________ 20.0 9.60×10^−2 ___ ___ _____________________20.0 ____________ 1.00 ______ 3.50×10^−2 _____25.0

0.0370 mol/(L⋅atm) The solubility of a gas in a liquid increases with increasing pressure. To understand the above statement, consider a familiar example: cola. In cola and other soft drinks, carbon dioxide gas remains dissolved in solution as long as the can or bottle remains pressurized. As soon as the lid is opened and pressure is released, the CO2 gas is much less soluble and escapes into the air. The relationship between pressure and the solubility of a gas is expressed by Henry's law: S=kP, where S is concentration in M, k is the Henry's law constant in units of mol/(L⋅atm), and P is the pressure in atm. Note: Since temperature also affects the solubility of a gas in an liquid, the Henry's law constant is specific to a particular gas at a particular temperature.

Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that freezes at −14.5 ∘C? The freezing point for pure water is 0.0 ∘C and Kf is equal to 1.86 ∘C/m.

108 g (Use ΔTf = Kf⋅m⋅i to find molality then use molality to find grams)

Glucose makes up about 0.11% by mass of human blood. Calculate this concentration in ppm. Express your answer in parts per million to two significant figures.

1100 ppm (0.11% is 0.11 out of 100 so create a ratio and solve for x: .11/100 = x/1,000,000)

https://session.masteringchemistry.com/problemAsset/3237512/1/BLB.ch14.p006.jpg Given the following diagrams at t = 0 and t = 30, what is the half-life of the reaction if it follows first-order kinetics?

15 min.

Consider the following energy profile. https://session.masteringchemistry.com/problemAsset/3237641/1/SAE_14_13_2048.png Which of the following statements is/are true? I. C and E are intermediates. II. B, D, and F are transition states. III. This is a three-step mechanism.

All three statements are true. If we compare this energy profile to the one shown in Figure 14.19, we can see that C and E are intermediates, B, D, and F are transition states, and the three steps of the mechanism, which are A → C, C → E, and E → G add to give the chemical equation of the overall process, which is A → G.

Yes.

https://session.masteringchemistry.com/problemAsset/1061737/43/BLB-1061737-graph.jpg For which salt, K2SO4, Li2SO4, KClO4, or NH4Cl, will increasing the temperature of the water have the greatest change in solubility per 100 g solvent

NH4Cl You can estimate the slopes qualitatively by looking to see which line is the steepest. You could also calculate values by finding Δy/Δx. If you do this, you should have calculated values approximately equal to the following: (Salt & Slope) Li2SO4 -0.07 K2SO4 0.13 KClO4 0.17 NH4Cl 0.47

The addition of NO accelerates the decomposition of N2O, possibly by the following mechanism: NO(g) + N2O(g) → N2(g) + NO2(g) 2NO2(g) → 2NO(g) + O2(g) Is NO serving as a catalyst or an intermediate in this reaction?

NO is a catalyst In the overall reaction, NO is introduced with the reactant but undergoes no permanent chemical change. It only increases the rate of decomposition of N2O to NO and O2, which makes it a catalyst.

If a molecule like Cl2 falls apart in an elementary reaction, what is the molecularity of the reaction? ____________. An elementary reaction occurs in ___________. The order of an elementary reaction is given by the coefficients in the chemical reaction. For the decomposition of Cl2, the coefficient of Cl2 is ___________.

*Unimolecular*. An elementary reaction occurs in *a single step*. The order of an elementary reaction is given by the coefficients in the chemical reaction. For the decomposition of Cl2, the coefficient of Cl2 is *one*.

A second-order reaction has a rate law: Rate = k[A]^2, where k = 0.150 M−1s−1. If the initial concentration of A is 0.250 M, what is the concentration of A after 5.00 minutes? Be sure to pay attention to the units in this problem so that they cancel out.

0.0204 M The integrated rate law for a second-order reaction is: 1/[A] = kt + 1/[A]0. If you got 49 as the answer you forgot that that is the answer for 1/[A] and so you have to do 1/49 to get the value of [A].

What pressure is required to achieve a CO2 concentration of 9.60×10^−2 M at 20∘C? S (mol/L) ____ P (atm) ___ k (mol/(L⋅atm) ___ T (∘C) 3.70×10^−2 __ 1.00 _____________________ 20.0 9.60×10^−2 ___ ___ _____________________20.0 ____________ 1.00 ______ 3.50×10^−2 _____25.0

2.59 atm (S is the concentration. S = kP (Henry's Law). You can find the value of k by using the 1st set of info. Don't use the value of k in the 3rd set because that is at 25.0C, not 20.0C and the value of k varies with temperature.)

https://session.masteringchemistry.com/problemAsset/3236338/1/Brown13.ch13.f1.jpg What aspect of the kinetic theory of gases tells us that the gases will mix? The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained. As long as temperature remains constant, the average kinetic energy of the molecules does not change with time. Energy can be transferred between molecules during collisions. Attractive and repulsive forces between gas molecules are negligible. Gases consist of large numbers of molecules that are in constant random motion. At any given temperature, the molecules of all gases have the same average kinetic energy.

Gases consist of large numbers of molecules that are in constant random motion.

Draw the Lewis structure for NOF. Nitrogen is the central atom.

N is double bonded to O and single bonded to F. O has 4 valence electrons, N has 2 valence electrons, and F has 6 valence electrons.

Consider the following elementary steps that make up the mechanism of a certain reaction: 2X→Y+Z Y+2L→M+Z Which species is a reaction intermediate?

Y Reaction intermediates are produced in one step and consumed in another. Thus they do not appear in the overall reaction. Therefore, you need to identify the species that cancels out and is not present in the overall reaction.

Yes.

After four half-life periods for a first-order reaction, what fraction of reactant remains?

[A]t = 6.25×10^−2 A0 (1/16 is also an acceptable answer)

Glucose makes up about 0.11% by mass of human blood. Calculate the molality. Express your answer in moles per kilogram of solvent to two significant figures.

6.1×10^−3m (.11% means for every 100 grams of blood, .11g is glucose and 99.89g is solvent. Find moles of solute (glucose) in .11g and divide it by kg of solvent.)

What is the mass percentage of iodine (I2) in a solution containing 3.0×10^−2 mol I2 in 105 g of CCl4?

6.8% (1st: The amount of grams of I2 in .03 mols must be found. mol = g/molar mass therefore (mols)(molar mass) = grams. (0.03 mols)(253.808 g/mol) = 7.61424 g. 2nd: Mass percentage = (mass of solute) / (total mass of solution) x (100%) or (mass of I2) / (mass of I2 + mass of CCl4) x (100%). 7.61424g / (7.61424g + 105g) x (100%) = 6.8%

Based on their available oxidation states, rank the following metals on their ability to catalyze this and other oxidation-reduction reactions. Rank from best to worst catalyst. To rank items as equivalent, overlap them. vanadium titanium manganese

Best catalyst: manganese Medium: vanadium Worst: titanium Manganese is especially useful because it can form every oxidation state from +2 to +7. Vanadium catalysts are most commonly used to produce sulfuric acid and titanium catalysts can be used to make various polymers. Catalysts used in oxidation-reduction reactions often transfer electrons between reactants in smaller steps than can be done in the uncatalyzed reaction. This means that the catalysts can form many oxidation states to take or give one or two electrons at a time. In other words, an element with many available oxidation states can be used as a catalyst for many oxidation-reduction reactions. Examine the given metal ions for their possible number of oxidation states. The number of possible oxidation states for a metal is often related to the number of electrons available in its outermost subshells.

Classify each process as an endothermic or exothermic process. forming solute-solvent attractions breaking solvent-solvent attractions breaking solute-solute attractions

Exothermic: forming solute-solvent attractions Endothermic: breaking solvent-solvent attractions breaking solute-solute attractions To dissolve one substance in another, thus forming a solution, three things must occur: The forces holding solute particles together must be broken. The forces holding solvent particles together must be broken. Attractions between solute and solvent particles must be formed.

https://session.masteringchemistry.com/problemAsset/3237493/4/MC_3237493_graph.png Does the instantaneous rate increase, decrease, or remain the same as the reaction proceeds?

It decreases

https://session.masteringchemistry.com/problemAsset/3236345/1/Figure_P13.9.jpg Where would you expect N2 to fit on this graph?

Its curve will be just below that of CO. N2 has the same molecular weight as CO but is nonpolar, so we can predict that its curve will be just below that of CO.

Consider the following reaction: O2+2NO→2NO2, rate = k[O2][NO]^2 What are the units of the rate constant k for this reaction?

M−2⋅s−1 The overall reaction order determines the units of the rate constant. Since the units of the rate are always M/s, the rate-constant units must account for s in the denominator as well as cancel out any extra M's that come from concentrations raised to an exponent. For the units of rate to come out to be M/s, the units of the rate constant for third-order reactions must be M−2⋅s−1 since M/s = (M−2⋅s−1)(M3) For a second-order reaction, the rate constant has units of M−1⋅s−1 because M/s=(M−1⋅s−1)(M2). In a first-order reaction, the rate constant has the units s−1 because M/s = (s−1)(M1).

Metals often form several cations with different charges. Cerium, for example, forms Ce3+ and Ce4+ ions, and thallium forms Tl+ and Tl3+ ions. Cerium and thallium ions react as follows: 2Ce^4+(aq) + Tl+(aq) → 2Ce^3+(aq) + Tl^3+(aq) If the uncatalyzed reaction occurs in a single elementary step, why is it a slow reaction? The reaction requires the collision of three particles with the correct energy. The transition state is low in energy. The probability of an effective three-particle collision is low. All reactions that occur in one step are slow.

The reaction requires the collision of three particles with the correct energy. The probability of an effective three-particle collision is low. Generally collisions between particles with the correct energy and orientation both must occur for a reaction to proceed. However, since the specific reagents are symmetrical cations, collisions will be at the correct orientation regardless.

https://session.masteringchemistry.com/problemAsset/1061737/43/BLB-1061737-graph.jpg A solid mixture consists of 40.8 g of KNO3 (potassium nitrate) and 7.2 g of K2SO4 (potassium sulfate). The mixture is added to 130. g of water. Consider the mixture described in the introduction. At 60 ∘C, to what extent will each solute be dissolved? (Totally, partially, or not dissolved) KNO3 K2SO4

Totally dissolved: KNO3 K2SO4 Partially dissolved: (None) Not dissolved: (None)

https://session.masteringchemistry.com/problemAsset/3237507/1/MC_2537914_graph.jpg Consider the following graph of the concentration of a substance X over time. Identify each of the following statements as true or false. The average rate between points 1 and 2 is greater than the average rate between points 1 and 3. As time progresses, the curve will eventually turn downward toward the x axis. X is a product of the reaction. The rate of the reaction remains the same as time progresses.

True: X is a product of the reaction. The average rate between points 1 and 2 is greater than the average rate between points 1 and 3. False: As time progresses, the curve will eventually turn downward toward the x axis. The rate of the reaction remains the same as time progresses. According to the graph, X is a product because its concentration increases over time. As the reaction proceeds, its rate decreases because the concentration of the reactants decreases. The average rate of a reaction between any two points on the graph is the slope of the line connecting the two points. Thus, the average rate of the formation of products between points 1 and 2 is greater than that between points 1 and 3. Finally, since the graph shows the formation of X with time, the concentration of X will not decrease once it reaches its maximum. Instead, it will reach a plateau.

Suppose that for the reaction K+L⟶M, you monitor the production of M over time, and then plot the following graph from your data: https://session.masteringchemistry.com/problemAsset/3237509/8/Brown13.ch14.p04.jpg Suppose the reaction as plotted here were started with 0.20 mol K and 0.40 mol L. After 30 min, an additional 0.20 mol L are added to the reaction mixture. Which of the following correctly describes how the plot would look from t=30min to t=60min? [M] would increase with the same slope as t = 0 to 15 min, until t = 45 min at which point the plot becomes horizontal again. [M] would remain at the same constant value it has at t=30min. [M] would decrease with the opposite slope to that at t=0 to 15 min, until t=45min at which point the plot becomes 0.

[M] would remain at the same constant value it has at t=30min. (L isn't the limiting reagent so it doesn't matter how much more you add, the reaction won't continue.)

https://session.masteringchemistry.com/problemAsset/3237492/1/Figure_P14.1.jpg Estimate the number of moles of A in the mixture after 30 s. between 0.46 and 0.58 mol 0.58 mol between 0.58 and 0.70 mol 0.41 mol between 0.30 and 0.41 mol between 0.41 and 0.54 mol

between 0.30 and 0.41 mol (Mastering changed the number from 0.42 in the book to 0.41) Our first guess might be half way between the values at 20s and 40s, namely 0.42 mol A. However, we also see that the change in the number of moles of A between 0s and 20s is greater than that between 20s and 40s -- in other words, the rate of conversion gets smaller as the amount of A decreases. So we would guess that the change from 20s to 30s is greater than the change from 30 s to 40 s, and we would estimate that the number of moles of A is between 0.42 and 0.30 mol.

When ammonium chloride dissolves in water, the solution becomes colder. Is the solution process exothermic or endothermic?

endothermic In an endothermic reaction, energy needs to be added to the system to form the products, which will have a greater enthalpy than the reactants. Since energy is absorbed by the system, the temperature of the surroundings decreases.

An unsaturated solution is one that has a concentration lower than the solubility has no double bonds contains no solute contains the maximum concentration of solute possible, and is in equilibrium with undissolved solute contains more dissolved solute than the solubility allows

has a concentration lower than the solubility

If the 3.90 m solution from Part A boils at 103.45 ∘C, what is the actual value of the van't Hoff factor, i? The boiling point of pure water is 100.00 ∘C and Kb is equal to 0.512 ∘C/m.

i = 1.73 (Use ΔTb = Kf⋅m⋅i to solve for i) Because of incomplete dissociation, 1 mol of KNO3 will produce a total of 1.73 mol of ions under these conditions.

Consider the reaction A + 2B ⇌ C whose rate at 25 ∘C was measured using three different sets of initial concentrations as listed in the following table: Trial ___ [A](M) ___ [B](M) ___ Rate(M/s) 1 ______ 0.40 ____ 0.050 ____ 9.6×10^−3 2 ______ 0.40 ____ 0.100 ____ 1.9×10^−2 3 ______ 0.80 ____ 0.050 ____ 3.8×10^−2 Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M

initial rate = 2.3×10^−2 M/s First, calculate the rate constant using the data from any trial and plugging the values into the rate law (Rate = k[A]^2[B]). Next, use the rate law and substitute the calculated rate constant and given concentrations to find the initial rate of the reaction. Be careful not to confuse the rate of a reaction and the rate constant. The rate depends on concentrations, whereas the rate constant does not (i.e., it is a constant). The rate always has units of M/s, whereas the units of the rate constant depend on the overall reaction order.

Consider the following multistep reaction: A+B→AB (slow) A+AB→A2B(fast) -------------------- 2A+B→A2B(overall) Based on this mechanism, determine the rate law for the overall reaction. Express your answer in standard MasteringChemistry format. For example, if the rate law is k[A]3[B]2 type k*[A]^3*[B]^2.

k[A][B] A multistep reaction can only occur as fast as its slowest step. Therefore, it is the rate law of the slow step that determines the rate law for the overall reaction.

A hypothetical solution forms between a solid and a liquid. The values of the thermodynamic quantities involved in the process are shown in the following table. separation of solute: 16.5 kJ/mol separation of solvent: 22.8 kJ/mol formation of solute-solvent interactions: -91.7 kJ/mol solute Calculate the enthalpy of solution in kilojoules per mole of solute.

ΔHsoln = -52.4 kJ/mol The enthalpy of the overall solution process is the sum of the component enthalpies as shown in the following equation: ΔHsoln=ΔH1+ΔH2+ΔH3 When using this equation, be aware of the signs on all enthalpy terms. If ΔHsoln is positive, the formation of the solution is endothermic. Endothermic dissolution reactions such as ammonium nitrate dissolving in water are used in the "instant ice packs" used to treat athletic injuries. The formation of the solution absorbs so much energy that the temperature of the "ice pack" decreases.

The rate constant of a chemical reaction increased from 0.100 s−1 to 2.90 s−1 upon raising the temperature from 25.0 ∘C to 53.0 ∘C . Calculate the value of (1/T2) − (1/T1) where T1 is the initial temperature and T2 is the final temperature.

−2.88×10^−4 K^−1 Remember to convert the temperatures to Kelvin.

For the reaction 2AsH3 (g) → As2 (g) + 3H2 (g), the reaction rate can be given by which of the following? +1/2∆[AsH3]/∆t +3∆[H2]/∆t +1/2∆[As2]/∆t +∆[As2]/∆t

+∆[As2]/∆t The reaction rate for the general reactions aA + bB → cC + dD is: Rate = −(1/a)∆[A]/∆t = (−1/b)∆[B]/∆t = (+1/c)∆[C]/∆t = (+1/d)∆[D]/∆t

The rate constant of a chemical reaction increased from 0.100 s−1 to 2.90 s−1 upon raising the temperature from 25.0 ∘C to 53.0 ∘C . Calculate the value of ln(k1/k2) where k1 and k2 correspond to the rate constants at the initial and the final temperatures as defined in part A.

-3.37

A sulfuric acid solution containing 571.4 g of H2SO4 per liter of aqueous solution has a density of 1.329 g/cm3. Calculate the mole fraction of H2SO4 in this solution. Express your answer to three significant figures.

0.122 The mole fraction of a component in the solution is the amount of the component in the solution divided by the total moles of all components: χcomponent = (moles of component) / (total moles of all components) Begin by calculating the mass of the solution using a set of conversion factors. As shown in Part A, the mass of the solution is 1329 g. The mass of the solution is the sum of the masses of the solute and the solvent. Subtract the given mass of sulfuric acid from the mass of the solution to calculate the mass of water contained in the solution: mass of H2O = 1329g − 571.4g = 758g Then, calculate the number of moles of each component of the solution using the molar masses of H2SO4 and H2O as conversion factors: moles of H2SO4 = 571.4g H2SO4 × (1 mol H2SO4 / 98.09g H2SO4) = 5.825 mol H2SO4 moles of H2O = 758g H2O × (1 mol H2O / 18.02g H2O) = 42.0 mol H2O Finally, substitute all the quantities into the equation for the mole fraction of a component: χH2SO4 = 5.825 mol / (5.825 mol + 42.0 mol) = 0.122

What is the molarity (M) of a solution prepared by dissolving 15.0 grams of Na3PO4 in enough water to make 250.0 mL of solution?

0.366M Molarity (M) = moles of solute/L of soln. 1st: Find how many moles of Na3PO4 is in 15.0 grams (it's 0.0914969 mols) 2nd: convert 250.0 mL to Liters (it's .2500 L) 3rd: Do mol/L (.0914969/.2500 = 0.366M)

1 The rate tripled when [C] was tripled. This indicates an order of 1 with respect to C because 3=3^1. Compare the values in trial 1 to those in trial 2. Notice that [A] and [B] remain constant whereas [C] triples. What happens to the rate in the transition from trial 1 to trial 2? It triples. The reaction rate can be determined from the following equation: rate = k[A]^x[B]^y[C]^z where z is the order with respect to C. For trial 1 this equation is: 3.0 × 10^−5 = k(0.10)^x(0.10)^y(0.10)^z and for trial 2 this equation is: 9.0 × 10^−5 = k(0.10)^x(0.10)^y(0.30)^z Dividing the equation for trial 2 by that for trial 1 gives (9.0×10^−5) / (3.0×10^−5) = (k/k)(0.10 / 0.10)^x(0.10 / 0.10)^y(0.30 / 0.10)^z which simplifies to 3 = (1)(1)^x(1)^y(3)^z Now, solve for x (This can be done by dividing the log of the left side by the log of the right side. You have 3 = 3^y. Take log(3) and divide it by log(3) to get the value of the exponent.).

The rate law for the reaction NH2 (g) + NO (g) → N2 (g) + H2O (g) was determined to be as follows: Rate = k[NH2][NO]. In an experiment at 1200K, where the initial concentration of NH2 was 1.00 × 10^−5 M and the initial concentration of NO was 1.00 × 10^−5 M, the reaction rate was measured to be 0.12 M/s. Calculate the value of the rate constant.

1.2 × 10^9 M−1s−1 The rate constant, k can be calculated by using: Rate/[NH2][NO].

A certain first-order reaction (A→products) has a rate constant of 3.00×10^−3 s−1 at 45 ∘C. How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration?

15.4 min First, calculate the value of t1/2 (the half-life) of the reaction. Then determine the number of half-lives that must pass for the concentration to drop from 100% to 6.25%. The half-life of a reaction, t1/2, is the time it takes for the reactant concentration [A] to decrease by half. For example, after one half-life the concentration falls from the initial concentration [A]0 to [A]0/2, after a second half-life to [A]0/4, after a third half-life to [A]0/8, and so on. For a first-order reaction, the half-life is constant. It depends only on the rate constant k and not on the reactant concentration. It is expressed as t1/2 = 0.693/k For a second-order reaction, the half-life depends on the rate constant and the concentration of the reactant and so is expressed as t1/2 = 1/(k[A]0)

Calculate the mass of ethylene glycol (C2H6O2) that must be added to 1.00 kg of ethanol (C2H5OH) to reduce its vapor pressure by 10.5 torr at 35 ∘C . The vapor pressure of pure ethanol at 35 ∘C is 1.00×10^2 torr. Express your answer in grams to two significant figures.

160g The vapor-pressure lowering (ΔP) is expressed as the mole fraction of the solute (χC2H6O2) multiplied by the vapor pressure of a pure solvent (P∘C2H5OH). Rearrange the expression to solve for χC2H6O2: χC2H6O2 = (ΔP / P∘C2H5OH) = (10.5 torr / 1.00×10^2 torr) = 0.105 The sum of the mole fractions of the solute (χC2H6O2) and solvent (χC2H5OH) is equal to 1, so: χC2H5OH = (1.000 − χC2H6O) = (21.000 − 0.105) = 0.895 After that, calculate the number of moles of the solvent (ethanol) from its mass given by converting it from kilograms to grams and then to the number of moles using the molar mass of ethanol: moles C2H5OH = 1.00 kg C2H5OH × (1000g C2H5OH / 1kg C2H5OH) × (1 mol C2H5OH / 46.07g C2H5OH) = 21.7 mol C2H5OH The mole fraction of the solvent calculated earlier (χC2H5OH) is the moles of the solvent divided by the total number of moles of the solute and solvent. Rearrange the expression of the mole fraction of the solvent to solve for the total number of moles: total moles = (mol C2H5OH / χC2H5OH) = (21.7 mol / 0.895) = 24.3mol Now find the moles of the solute from the total number of moles in the solution and the moles of the solvent: mol C2H6O2 = (total moles − mol C2H5OH) = (24.3mol − 21.7 mol) = 2.5mol Finally, convert the moles of ethylene glycol to the mass of ethylene glycol in grams using its molar mass (62.07 g/mol) as a conversion factor: mass C2H6O2 = 2.5 molC2H6O2 × (62.07g C2H6O2 / 1 mol C2H6O2) = 160g C2H6O2

2 The rate quadrupled when [A] was doubled. This indicates an order of 2 with respect to A because 4 = 2^2. Compare the values in trial 1 to those in trial 3. Notice that [B] and [C] remain constant whereas [A] doubles. What happens to the rate in the transition from trial 1 to trial 3? It Quadruples. The reaction rate can be determined from the following equation: rate=k[A]^x[B]^y[C]^z where x is the order with respect to A. For trial 1 this equation is: 3.0 × 10^−5 = k(0.10)^x(0.10)^y(0.10)^z and for trial 3 this equation is: 1.2×10^−4 = k(0.20)^x(0.10)^y(0.10)^z Dividing the equation for trial 3 by that for trial 1 gives (1.2×10^−4) / (3.0×10^−5) = (k/k)(0.20 / 0.10)^x(0.10 / 0.10)^y(0.10 / 0.10)^z which simplifies to 4 = (1)(2)^x(1)^y(1)^z Now, solve for x (This can be done by dividing the log of the left side by the log of the right side. You have 4 = 2^x. Take log(4) and divide it by log(2) to get the value of the exponent.).

2.2×10^−3 M/s

Paradichlorobenzene, C6H4Cl2, is a component of mothballs. A solution of 2.00 g in 22.5 g of cyclohexane boils at 82.39 ∘C. The boiling point of pure cyclohexane is 80.70 ∘C. Calculate Kb for cyclohexane.

2.80∘C/m (Mastering accepted 2.79) The temperature at which a solution freezes and boils depends on the freezing and boiling points of the pure solvent as well as on the molal concentration of particles (molecules and ions) in the solution. For nonvolatile solutes, the boiling point of the solution is higher than that of the pure solvent and the freezing point is lower. The change in the boiling for a solution, ΔTb, can be calculated as ΔTb=Kb⋅m in which m is the molality of the solution and Kb is the molal boiling-point-elevation constant for the solvent. The freezing-point depression, ΔTf, can be calculated in a similar manner: ΔTf=Kf⋅m in which m is the molality of the solution and Kf is the molal freezing-point-depression constant for the solvent.

21 g (The solubility of KNO3 at 0C is about 15g/100g of water. The KNO3 is added to 130g of water. 15g of KNO3/100g of water X 130 g of water is 19.5 g of KNO3 is capable of dissolving at 0C. All of the KNO3 above 19.5g crystallizes so 40.8 - 19.5 = 21.3g. Because 2 sig figs were used for the 15g value 21.3 gets rounded down to 21 g of KNO3 that crystalizes.)

Calculate the vapor pressure of water above a solution prepared by dissolving 27.5 g of glycerin (C3H8O3) in 143 g of water at 343 K. (The vapor pressure of water at 343 K is 233.7 torr.) Express your answer in torrs to three significant figures.

225 torr The vapor pressure of a solution is lower than that of a pure solvent at the same temperature, and it decreases in proportion to the mole fraction for the solvent. First, calculate the number of moles of water and glycerin from the mass given (in grams) by using the molar mass of water (18.02 g/mol) and glycerin (92.09 g/mol) as conversion factors: mol H2O = 143 gH2O × (1 mol H2O / 18.02g H2O) = 7.94 mol H2O mol C3H8O3 = 27.5 g C3H8O3 × (1 mol C3H8O3 / 92.09g C3H8O3) = 0.299 mol C3H8O3 Then calculate the mole fraction of water (χH2O) by dividing the number of moles of water by the total number of moles of water and glycerin: χH2O = mol H2O / (mol H2O + mol C3H8O3) = 7.94mol / (7.94 mol + 0.299 mol) = 0.964 Finally, use the mole fraction of water and the vapor pressure of pure water (P∘H2O) at 343 K (233.7 torr) to determine the vapor pressure of water above the solution (PH2O) by Raoult's law: PH2O = (χH2O)(P∘H2O) = 0.964 × 233.7torr = 225torr

Calculate the solubility (in M units) of ammonia gas in water at 298 K and a partial pressure of 4.50 bar . The Henry's law constant for ammonia gas at 298 K is 58.0 M/atm and 1 bar=0.9869 atm.

258 M According to Henry's law, the solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas over the solution. This is expressed by a simple chemical equation, S = k × P where S is the solubility of the gas in the liquid, P is the partial pressure of the gas over the liquid, and k is Henry's law constant. Henry's law constant has units of M/atm. Thus, the units for pressure must be atm. Convert the units of pressure from bar to atm, using the relation 1bar=0.9869atm, and then substitute the values into the equation expressing Henry's law. Using Henry's law equation, the solubility of ammonia gas is calculated as S = k × P = 58M/atm × 4.44atm = 258M The solubility of gas in a liquid solvent depends upon the partial pressure of the gas over the solvent, temperature, and the specific type of solvent.

The addition of NO accelerates the decomposition of N2O, possibly by the following mechanism: NO(g) + N2O(g) → N2(g) + NO2(g) 2NO2(g) → 2NO(g) + O2(g) Consider how the two steps can be added to give the overall equation, and then write the chemical equation for the overall reaction. Express your answer as a chemical equation including phases.

2N2O(g)→2N2(g)+O2(g) NO accelerates the decomposition of N2O and therefore is not a reactant or product itself. The steps should sum such that NO is canceled from the overall equation by multiplying all coefficients of the first step by 2. 2NO(g) + 2N2O(g) → 2N2(g) + 2NO2(g) 2NO2(g) → 2NO(g) + O2(g) ---------------------------------------------- 2N2O(g) → 2N2(g) +O2(g)

Consider the following elementary steps that make up the mechanism of a certain reaction: 2X→Y+Z Y+2L→M+Z What is the overall reaction?

2X+2L→2Z+M To add two reactions together, first write all the species on one line. Then cancel out anything that appears on both sides of the arrow. For the given reaction, you would have 2X+Y+2L→Y+Z+M+Z

Consider the following reaction: O2+2NO→2NO2, rate = k[O2][NO]^2 What is the overall reaction order?

3 The overall reaction order is the sum of the reaction orders with respect to each reactant. For the general rate law k[A]m[B]n, the overall reaction order is m+n. The sum of the individual orders gives the overall reaction order. The example equation, k[A]^0[B]^1[C]^2, is third order overall because 0+1+2=3.

A sulfuric acid solution containing 571.4 g of H2SO4 per liter of aqueous solution has a density of 1.329 g/cm3. Calculate the mass percentage of H2SO4 in this solution. Express your answer in percent to four significant figures.

42.99 % The mass percentage of a component in the solution is the mass of the component in the solution divided by the mass of that solution all multiplied by 100%: (mass % of component) = (mass of component in solution / total mass of solution) × 100% Begin by calculating the mass of the solution using a set of conversion factors. Since density has the units of g/cm3, use the appropriate conversion factor to get the volume of the solution in cm3: mass of solution = 1L × (1000cm3 / 1L) × (1.329g / 1cm3) = 1,329g Substitute all the quantities into the equation for the mass percentage of a component: mass % of H2SO4 = (571.4g / 1329g) × 100% = 42.99%

The reaction of hydrogen peroxide with iodine, H2O2(aq) + I2(aq) ⇌ OH−(aq) + HIO(aq) is first order in H2O2 and first order in I2. If the concentration of H2O2 was increased by half and the concentration of I2 was increased by four, by what factor would the reaction rate increase?

6 For simplicity, it will help to express the initial concentrations of H2O2 and I2 as x and y, respectively. You need to first determine how the initial rate of the reaction can be expressed in terms of x and y. Then determine the new concentrations of the reactants in terms of x and y. For example, if the concentration of H2O2 was increased by one-fourth, then the new concentration of H2O2 in terms of x will be: x + 1/4x = 1x + 0.25x = 1.25x Similarly, if the concentration of H2O2 was doubled, then the new concentration of H2O2 in terms of x will be: x + x = 2x If you substitute the new concentrations of the reactants in the rate-law equation, you will get the new rate of the reaction. Then divide the new rate by the original rate to see by how much it increased. If the initial concentrations are [H2O2] = x and [I2] = y, the reaction rate is: rate = kxy the new rate would be: new rate = k(1.5x)(4y) = 6kxy 6kxy / kxy = 6

A sulfuric acid solution containing 571.4 g of H2SO4 per liter of aqueous solution has a density of 1.329 g/cm3. Calculate the molality of H2SO4 in this solution. Express your answer in moles per kilogram to four significant figures.

7.689 mol/kg (Mastering accepted my answer of 7.690) The molality of a solution is the number of moles of solute per kilogram of solvent: molality = (moles of solute) / (kilograms of solvent) Begin by calculating the mass of the solution using a set of conversion factors. As shown in Part A, the mass of the solution is 1329 g. The mass of the solution is the sum of the masses of the solute and the solvent. Subtract the given mass of sulfuric acid from the mass of the solution to calculate the mass of water contained in the solution and convert it to kg using the appropriate conversion factor: mass of H2O = (1329g − 571.4g) × (1kg / 1000g) = 0.758kg Calculate the amount of moles of H2SO4 using its molar mass as a conversion factor. As shown in Part B, there are 5.825 mol of H2SO4. Finally, substitute all the quantities into the equation for the molality of a solution: molality of H2SO4 = (5.825mol / 0.758kg) = 7.69m

Concentration of aqueous ammonia is 15 % NH3 by mass and solution has a density of 0.94 g/mL. What is the molarity of this solution? Express the molarity to two significant figures.

8.3 M To find the molarity of the solution, determine the mass of 1.0 L of the ammonia solution first. To do this, use the density of the entire solution to convert its volume to mass in grams. Recall that 1 L=1000 mL. mass of NH3 solution = 1L solution × (1000mL solution / 1L solution) × (0.94 g solution / 1mL solution) = 940 g solution The mass percentage of a component is the ratio between the mass of that component and the total mass of the solution multiplied by 100%. Therefore, 15%m of the total mass of the given solution can be calculated as follows: mass of NH3 = g solution × (g NH3 / 100 g solution) = 940g solution × (15 g NH3 / 100g solution) = 140 g NH3 This is the mass of ammonia in one liter of solution. Molarity is the number of moles of a solute per liter of solution, so to find the molarity, divide the mass of ammonia in one liter of the solution by its molar mass: moles of NH3 per liter = 140g NH3 × (1 mol NH3 / 17.03g NH3) = 8.3 mol NH3 Since this is the number of moles of ammonia in one liter of solution, it is also the molarity of the solution.

For the hypothetical reaction A → B, calculate the average rate of disappearance of A if the initial concentration of A is 0.91 M and the concentration of A after 90 minutes is 0.11 M.

8.9 × 10^−3 M/min

The rate constant of a chemical reaction increased from 0.100 s−1 to 2.90 s−1 upon raising the temperature from 25.0 ∘C to 53.0 ∘C . What is the activation energy of the reaction? Express your answer numerically in kilojoules per mole.

97.1 kJ/mol This large Ea value is the reason why a minor increase in temperature caused the reaction to proceed about 29 times faster. Reactions with a small energy barrier do not see such a drastic increase in rate as temperature increases. As temperature rises, the average kinetic energy of molecules increases. In a chemical reaction, this means that a higher percentage of the molecules possess the required activation energy, and the reaction goes faster. This relationship is shown by the Arrhenius equation: k=Ae^(−Ea/RT) where k is the rate constant, A is the frequency factor, Ea is the activation energy, R = 8.3145 J/(K⋅mol) is the gas constant, and T is the Kelvin temperature. The following rearranged version of the equation is also useful: ln(k1/k2) = (Ea/R)(1/T2 − 1/T1) where k1 is the rate constant at temperature T1, and k2 is the rate constant at temperature T2.

How does the presence of a catalyst affect the activation energy of a reaction? A catalyst increases the activation energy of a reaction. A catalyst decreases the activation energy of a reaction. A catalyst does not affect the activation energy of a reaction. It depends on whether you are talking about the forward or the reverse reaction.

A catalyst *decreases* the activation energy of a reaction. It is important to realize that the activated complex in the catalyzed reaction is not the same activated complex as in the uncatalyzed reaction. The activation energy is lowered because the catalyst provides an alternate path for the reactants to get to products. In other words, the catalyst provides an alternate activated complex.

How does the presence of a catalyst affect the enthalpy change of a reaction? https://session.masteringchemistry.com/problemAsset/1244768/11/uncatalyzed.jpg https://session.masteringchemistry.com/problemAsset/1244768/11/catalyzed_V2.jpg A catalyst increases the enthalpy change of a reaction. A catalyst decreases the enthalpy change of a reaction. A catalyst does not affect the enthalpy change of a reaction. It depends on whether you are talking about the forward or the reverse reaction.

A catalyst does not affect the enthalpy change of a reaction. This question illustrates the meaning of a state function. The enthalpy change of a reaction is the same whether it takes the catalyzed or uncatalyzed path. ΔH, like all state functions, only depends on the starting and ending points of the system.

https://media.pearsoncmg.com/bc/bc_0media_chem/legacy_media/int_act_hen_law/index.html Click on the Run button in the activity. Watch how the solubility of a gas changes as pressure is increased and then decreased during the run. Which of the following statements correctly explain the relationship between the solubility of a gas and its pressure? A. Fewer gas molecules are soluble at higher pressures. B. The solubility of a gas decreases with a decrease in pressure. C. More gas molecules are soluble as pressure is increased. D. As pressure is increased, the rate at which gas molecules enter the solution decreases. E. As pressure decreases, the concentration of gas molecules in the solution increases. F. The solubility of nitrogen gas at 2.00 atm is twice the solubility of the gas at 1.00 atm. G. The concentration of gas particles in the solution is higher at 4.25 atm than at 1.00 atm.

B,C,F, & G B. The solubility of a gas decreases with a decrease in pressure. C. More gas molecules are soluble as pressure is increased. F. The solubility of nitrogen gas at 2.00 atm is twice the solubility of the gas at 1.00 atm. G. The concentration of gas particles in the solution is higher at 4.25 atm than at 1.00 atm.

You are in the kitchen. You have a party later today and want to make some strawberry gelatin. Your mom comes home from a long day at work and hands you two bottles. They both contain a clear, slightly viscous liquid. One has a label "A" on it and the other has a label "B" on it. Your mom says "One of these bottles contains a solution of purified collagen in water. The other contains glycerol and poison. I can't remember which is which. I need to go take a nap. I know you can figure out which is which though, honey. Don't drink the poison please." Being a good chemist from birth, you look at this as another fun challenge. First, you pour the liquid from each bottle into separate beakers. Then you think about it for a few minutes and decide what to do. Of the following four choices, which is the most reasonable path for you, given your constraints? Cool both beakers on ice while stirring. Take the one that turns into gelatin and heat it with sugar and strawberry juice concentrate. Boil both beakers while stirring then cool them on ice. Look for the one that turns into gel. Reboil that one with sugar and strawberry juice concentrate. Mix the two beakers together. Add sugar and strawberry juice concentrate and boil. Let it cool into gelatin. Add aluminum sulfate to both beakers. See which one precipitates. Add sugar and strawberry juice concentrate to that one and boil it to make gelatin.

Boil both beakers while stirring then cool them on ice. Look for the one that turns into gel. Reboil that one with sugar and strawberry juice concentrate. Thomas Graham studied colloids in the 1860s. He coined the terms sol and gel to describe the different forms of colloids. A sol is a liquid form of a colloid and a gel is the gelatinous form of a colloid. Some sols can be converted into gels by careful manipulation of the temperature or pH conditions. For example, an aqueous sol of collagen can be boiled and when it cools, it turns into a gel, sometimes called gelatin. The boiling denatures the collagen protein, exposing hydrophobic parts of the protein that will form intermolecular aggregates.

Metals often form several cations with different charges. Cerium, for example, forms Ce3+ and Ce4+ ions, and thallium forms Tl+ and Tl3+ ions. Cerium and thallium ions react as follows: 2Ce^4+(aq) + Tl+(aq) → 2Ce^3+(aq) + Tl^3+(aq) This reaction is very slow and is thought to occur in a single elementary step. The reaction is catalyzed by the addition of Mn2+(aq) according to the following mechanism: Ce^4+(aq) + Mn^2+(aq) → Ce^3+(aq) + Mn^3+(aq) Ce^4+(aq) + Mn^3+(aq) → Ce^3+(aq) + Mn^4+(aq) Mn^4+(aq) + Tl+(aq) → Mn^2+(aq) + Tl^3+(aq) The catalyzed reaction is first order in [Ce^4+] and first order in [Mn^2+]. Which of the steps in the catalyzed mechanism is rate determining?

Ce^4+(aq) + Mn^2+(aq) → Ce^3+(aq) + Mn^3+(aq) In this reaction, the first step is the bottleneck for the rate of the reaction. This means that as soon as the products of the first step are made they are quickly used up in the following two steps. The rate of the catalyzed reaction is faster than the rate of the uncatalyzed reaction because the rate depends on only two molecules colliding with the correct energy and orientation, so the probability of reactive collisions is higher in the catalyzed reaction. The rate-determining step determines the rate law of a multistep reaction. The order of reaction tells us the reactants and their coefficients in the rate-determining step. For example, if the order of reaction is found to be ath order in A and bth order in B, the rate-determining step must be of the form: aA+bB→products where the lowercase letters are the coefficients and the uppercase letters are the molecular formulas. In multistep reactions in which any intermediates are included in the rate law, they will ultimately be expressed in terms of initial reactants.

You have studied the gas-phase oxidation of HBr by O2: 4HBr(g)+O2(g)→2H2O(g)+2Br2(g) You find the reaction to be first order with respect to HBr and first order with respect to O2. You propose the following mechanism: HBr(g) + O2(g) → HOOBr(g) HOOBr(g) +HBr(g) → 2HOBr(g) HOBr(g) + HBr(g) → H2O(g) + Br2(g). If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?

Doesn't disprove

It is determined experimentally that the reaction E + F → X + Y is second order with respect to E and first order with respect to F. Which of the following statements is correct? Halving the concentration of E will decrease the reaction rate by half. Doubling the concentration of E will have no effect on the rate. Doubling the concentration of E will increase the reaction rate by a factor of four. Doubling the concentration of F will cause the reaction rate to increase by a factor of four.

Doubling the concentration of *E* will increase the reaction rate by a factor of *four*. The rate law for this reaction is: Rate = k[E]2[F].

Which of the following statements reasonably explains why this reaction has a low activation energy? (Regarding the reaction of NO + F2 --> NOF + F) The electron-deficient NO molecule is attracted to the electron-rich F2 molecule, so the driving force for the formation of the transition state is greater than simple random collisions. The formal positive charge on the NO molecule is attracted to the formal negative charge on the F2 molecule, so the driving force for the formation of the transition state is greater than simple random collisions. Because this reaction is only bimolecular, the frequency of random collision is very high. The small NO and F2 molecules have little steric hindrance; therefore random collisions occur more frequently. Due to the type of bonds being broken and formed in the transition state the stability of this transient structure is relatively high.

Due to the type of bonds being broken and formed in the transition state the stability of this transient structure is relatively high. What factor(s) determine the magnitude of the activation energy? Relative energy of the reaction transition state

For the elementary process 2NOCl(g)→Cl2(g)+2NO(g) the activation energy (Ea) and overall ΔE are 97.8 and 74.5 kJ/mol, respectively. What is the activation energy for the reverse reaction?

Ea(reverse) = 23kJ/mol The activation energy for the reaction is equal to the difference between the energy of reactants and the energy of the activated state (the highest energy point). Since the products of the forward reaction (Cl2(g)+NO(g)) have higher potential energy than the reactant (ΔE is positive), the forward reaction is endothermic. Given that the products of the forward reaction are the reactants in the reverse reaction, the reverse reaction is exothermic and has a lower energy barrier than the forward reaction. Therefore, Ea (reverse) would be equal to Ea for the forward reaction minus ΔE of the forward reaction: Ea(reversereaction) = Ea −ΔE = 97.8kJ/mol − 74.5kJ/mol = 23kJ/mol

Classify the steps involved in the formation of a solution as being endothermic or exothermic. separation of solute molecules separation of solvent molecules formation of solute-solvent interactions

Endothermic: separation of solute molecules separation of solvent molecules Exothermic: formation of solute-solvent interactions Exothermic processes release energy, whereas endothermic processes require energy. Think about solute and solvent particles as magnets. Energy is required to pull them apart, and energy is released when they are allowed to snap back together. The balance between the endothermic nature of the homogenous (solvent-solvent and solute-solute) interactions and exothermic nature of the heterogeneous (solute-solvent) interactions determines whether the overall enthalpy of solution is exothermic or endothermic. A solution is formed when the solute uniformly disperses throughout (or dissolves in) the solvent. The process can be described though three steps: separation of solvent-solvent particles, breaking of solute-solute particles, and formation of solute-solvent interactions. The overall energy change for the solution process, ΔHsoln , is the sum of the enthalpies of the three steps. Whether ΔHsoln is endothermic or exothermic depends on the relative magnitudes of ΔH1, ΔH2, and ΔH3, where the subscript indicates the step in the process corresponding to the enthalpy value.

https://media.pearsoncmg.com/bc/bc_0media_chem/legacy_media/int_act_ener_solu/index.html Predict what must happen when a beaker containing a solution of lithium chloride is heated. Check all that apply. Energy will be transferred into the system. The ΔH3 value will increase. The ΔHsoln value will decrease. The reaction will become endothermic.

Energy will be transferred into the system. Since ΔH values are constant for a given set of reaction conditions, they won't change when energy is added to or removed from the system. The equilibrium may change, that is, solute particles may dissolve more quickly or more slowly, but the ΔH values associated with each step won't change. When a beaker is heated, what direction is energy moving? Energy is entering the system when a beaker is heated. Since the dissolution of lithium chloride is exothermic, meaning that heat is released, adding more heat to the system does not increase the solubility of lithium chloride significantly.

Determine the enthalpy of solution for a solid with ΔH values as described in each scenario. a solid with ΔH values for the breaking of attractions that are one third the magnitude compared to forming of attractions a solid with ΔH values for the breaking of attractions that are three times the magnitude compared to forming of attractions a solid with ΔH values of approximately equal magnitude for each of the steps involved in the solution formation process

Exothermic: a solid with ΔH values for the breaking of attractions that are one third the magnitude compared to forming of attractions Endothermic: a solid with ΔH values for the breaking of attractions that are three times the magnitude compared to forming of attractions a solid with ΔH values of approximately equal magnitude for each of the steps involved in the solution formation process Approximately no change in energy: (None) The formation of a solution can be described in three discrete steps: the breaking of solute-solute attractions, the breaking of solvent-solvent attractions, and the forming of solute-solvent attractions. Therefore, the enthalpy of solution, ΔHsoln, is the sum of the enthalpy changes in the three steps: ΔHsoln=ΔH1+ΔH2+ΔH3 Try taking a values, such as 6 kJ/mol, and placing it in the formula to calculate the enthalpy of solution. Adjust the sign before the value taking into account whether the step is endothermic or exothermic. The sign of the answer will indicate the direction of energy flow. The dissolution of many salts, such as table salt, is slightly endothermic. The process of dissolving table salt, NaCl, in water has an enthalpy value of ΔHsoln=+4 kJ/mol. Therefore the temperature of the water will decrease slightly when the salt is added.

You have studied the gas-phase oxidation of HBr by O2: 4HBr(g)+O2(g)→2H2O(g)+2Br2(g) You find the reaction to be first order with respect to HBr and first order with respect to O2. You propose the following mechanism: HBr(g) + O2(g) → HOOBr(g) HOOBr(g) +HBr(g) → 2HOBr(g) HOBr(g) + HBr(g) → H2O(g) + Br2(g). Based on the rate law, which step is rate determining?

First (The rate law is rate = k[HBr][O2] and is determined by the rate determining step)

https://media.pearsoncmg.com/bc/bc_0media_chem/legacy_media/int_act_dis_ion/index.html Watch the animation in the activity and rank the events in the order that they occur as equilibrium is reached. Rank from the first step to the last step. To rank the steps that occur simultaneously, overlap them. hydration of cations hydrated cations and anions lose waters of hydration and deposit as the solid salt hydration of anions rate of dissolution is equal to the rate of recrystallization dissolution of salt into its cations and anions

First step: hydration of cations hydration of anions dissolution of salt into its cations and anions Second step: hydrated cations and anions lose waters of hydration and deposit as the solid salt Last Step: rate of dissolution is equal to the rate of recrystallization When salt is added to water, the solid salt dissociates into corresponding cations and anions, and, simultaneously, each of the ions is solvated by the solvent molecules. Since the solvent molecules are water, the process is known as hydration. You can observe in the animation that the dissolution and the hydration processes occur simultaneously, and the salt dissolves in the solvent (water). Initially, the dissolution of salt occurs rapidly. This results in a decrease in the amount of solid salt and an increase in the concentration of the dissolved salt in the solution. Gradually, some of the anions and cations begin to recrystallize or deposit. When the rate of dissolution equals the rate of recrystallization (deposition), equilibrium is attained. This equilibrium is said to be a dynamic equilibrium, as the process of the dissolution of salt and the process of recrystallization that forms solid NaCl occur at the same rate. At this point, the maximum possible amount of salt is dissolved, and the solution is said to be saturated. The interaction between the solute and the solvent plays an important role in the dissolution process. The force of attraction that can be formed between two substances determines their relative solubility. Substances with similar intermolecular forces are observed to be soluble in one another. An ionic substance will be more soluble in a polar solvent than in a nonpolar solvent. Similarly, a nonpolar compound will be soluble in a nonpolar solvent and is found to be insoluble in a polar solvent. To summarize all this in one statement it is often said that "Like dissolves like."

Rank the following substances in order from most soluble in water to least soluble in water: methane, CH4; 1-hexanol, C6H13OH; lithium nitrate, LiNO3; propane, C3H8

From most soluble to least soluble lithium nitrate > 1-hexanol > propane > methane The presence of London dispersion forces *increases the relative hydrophilicity of a gas*, however this trend is not observed between liquids. In water the liquid pentane, C5H12, is more soluble than the liquid hexane, C6H14. *For liquids, solubility in water often falls as the length of the non-polar hydrocarbon chain increases*. Chemists generally refer to solvents as hydrophilic (water-loving) or hydrophobic (water-hating) based on their ability to dissolve in water. The following factors, listed in order of importance, favor hydrophilicity: the presence of ions, the presence of a dipole moment, and the presence of London dispersion forces. The presence of ions: For example, a salt such as magnesium sulfate, MgSO4, is very soluble in water because the water molecules will solvate the charged magnesium atom and the sulfate molecule very easily. The presence of a dipole moment: A polar molecule will interact with water molecules, which also have a dipole moment. Dipole moments are caused by the close proximity of electronegative atoms such as oxygen, or nitrogen to less electronegative atoms such as hydrogen or carbon. Acetone, H3C(CO)CH3, and methanol, HOCH3, are readily soluble in water because of their dipole moments from the presence of the electronegative oxygen atom. In contrast, a molecule such as butane, CH3CH2CH2CH3, which has no dipole moment, is hydrophobic and hence insoluble in water. Hydrogen bonding is a strong dipole-dipole interaction that occurs when a hydrogen atom is bonded to a highly electronegative atom such as oxygen, nitrogen, or fluorine. The presence of hydrogen bonding will also increase solubility. The presence of London dispersion forces: Substances that lack ions or a dipole moment are generally hydrophobic. However, when comparing two nonpolar *gases*, the one with the greater London forces will generally be the more hydrophilic of the two. In general, London forces become larger as the size of the atom or molecule increases. So O2 (MW=32) is more soluble in water than N2 (MW=28).

HOOBr and HOBr (Intermediates are created in one step then used in another and so do not appear in the final equation)

List the following aqueous solutions in order of decreasing boiling point. Ignore ion pairing in your analysis. 0.050m Zn(NO3)2 0.120m glucose 0.050m LiBr

Highest boiling point: 0.050m Zn(NO3)2 Middle: 0.120m glucose Lowest boiling point: 0.050m LiBr (When it says to ignore ion pairing it means assume the ions completely dissociate, i.e. the ideal van't Hoff factor) Colligative properties depend on the total number of solute particles, where the type of solute particles can be ignored. The boiling points of these aqueous solutions can then be compared based on the concentrations of each solution and the dissociation of each solute. Since increasing solute amounts raises the boiling point, the solution with the highest boiling point also has the highest solute concentration. Zn(NO3)2 dissociates into three ions in solution, and the concentration of ions for the 0.050 m solution is expected to be greater than the 0.120 m concentration listed for glucose. Glucose is not an electrolyte and will not form ions in solution.

The progress of a reaction is determined by various steps involving energy changes with respect to collision theory. Identify the key features of the reaction profile below. Drag the appropriate labels to their respective targets in the energy profile diagram.

In the order of following the curves: Reactants, Ea1, Transition State 1 (at top of curve 1), Intermediate, Transition State 2 (at top of curve 2), Ea2, ΔH, Products A reaction profile is a graphical representation of reactants acquiring energy of activation to reach a hypothetical state called a transition state. The transition state decomposes to form a product. In two-step reactions, an intermediate is formed that is meta-stable; it then further undergoes a chemical change to attain a transition state that decomposes into products. A reaction profile is also known as an energy profile diagram as it gives an idea about the energy associated with reactants, products, and intermediate steps involved. In multistep reactions there are many transition states. The low-energy state between the consecutive transition states is called a reaction intermediate. It is a meta-stable state. A transition state is a state where the total kinetic energy of reactants is greater than the average kinetic energy that is sufficient to break bonds. In this state the breaking and forming of bonds are taking place simultaneously. Activation energy is the minimum energy that the molecules should attain so that the molecular collisions result in the formation of products. Enthalpy of the reaction is the difference between the total enthalpy of the reactants, ∑HR, and that of the products, ∑HP. Identify the relation between the activation energy, Ea, the total enthalpy of reactants, ∑HR, the total enthalpy of products, ∑HP, and the total enthalpy of the transition state, ∑HT. (Ea=∑HT−∑HR)

NaCl is an ionic solid. The Na+ and Cl− ions in NaCl are bonded through an electrostatic force of attraction commonly known as the ionic bond. Water is a polar solvent. The oxygen atom, being more electronegative, attracts the electron cloud toward itself. As the electron cloud is pulled by the oxygen atom, it carries a partial negative charge, and the hydrogen atoms carry a partial positive charge. This partial separation of charges in the water molecule makes it polar. Predict which intermolecular forces contribute the most to the dissolution of NaCl in water. Check all that apply. Dipole-dipole forces Ion-dipole forces Hydrogen bonding Ion-ion forces

Ion-dipole forces To predict the intermolecular forces that contribute the most to the dissolution of a substance in water, consider the force of attraction that can be formed between the substance and the dipoles of water molecules. The force of attraction that the substance experiences with water depends on whether it forms ions, has a hydrogen atom that can form hydrogen bonds, or has a dipole. For example, if the substance has a dipole, then the intermolecular force that contributes the most to the dissolution of the substance in water is predicted to be the dipole-dipole force. NaCl is an ionic solid. Na+ and Cl− ions are held together by an electrostatic force known as an ion-ion force. A water molecule is polar, and it shows hydrogen bonding. However, the dissolution of NaCl in water occurs as a result of the interaction of Na+ and Cl− with water dipoles. The negative dipole of water attracts Na+ ions and the positive dipole attracts Cl− ions. Thus, an ion-dipole intermolecular force exists between NaCl and water molecules. Ion-dipole forces play an important role in the dissolution of NaCl in water.

Suppose that NaCl is added to hexane (C6H14) instead of water. Which of the following intermolecular forces will exist in the system? Check all that apply. Ion-dipole force between Na+ ions and a hexane molecule Dipole-dipole force between two hexane molecules Ion-ion force between Na+ and Cl− ions Hydrogen bonding between Na+ ions and a hexane molecule London dispersion force between two hexane molecules

Ion-ion force between Na+ and Cl− ions London dispersion force between two hexane molecules To identify the intermolecular forces that exist in the system when NaCl is added to hexane, consider the intermolecular force between the ions of NaCl, the intermolecular force between the molecules of hexane, and the intermolecular force between ions of NaCl and the hexane molecule, if any. The intermolecular force between the molecules of hexane depends on the polarity of the hexane molecule. Solutions form when the magnitude of the attractive forces between solute and solvent particles is comparable to or greater than those that exist between the solute particles themselves or between the solvent particles themselves. The intermolecular force between Na+ and Cl+ ions is different than the intermolecular forces between hexane molecules. There is no solute-solvent interaction between NaCl and hexane; hence, NaCl is insoluble when the solvent is hexane.

Can an intermediate appear as a reactant in the first step of a reaction mechanism? It could be a ________ in the first step of a reaction mechanism, but not a ___________. reactant product catalyst

It could be a *product* in the first step of a reaction mechanism, but not a *reactant*.

A sealed beaker of what you are told is aqueous nickel sulfide was given to you by the local chemist and you put it on the window sill. After a few days, you notice that the mixture is getting thick toward the bottom. What is the nature of the initial nickel sulfide mixture, assuming that the temperature on the sill remained approximately constant. It is a solution. It is a suspension. It is a colloid. It is a gas.

It is a suspension A suspension is a mixture of materials that do not dissolve in each other, but one substance is suspended in the other. In this example, particles of insoluble nickel sulfide are fine enough to be suspended in water, but eventually settle over time. Colloids would not be expected to settle over time.

Appropriate units for the speed of a chemical reaction, the reaction rate, are g/mol M/s sec min

M/s

Consider the formation of the three solutions shown in the table. Solution A: solute-solute: weak solvent-solvent: weak solute-solvent: strong Solution B: solute-solute: weak solvent-solvent: strong solute-solvent: strong Solution C: solute-solute: strong solvent-solvent: strong solute-solvent: weak Rank the formation of the solutions A, B, and C from the most exothermic to the most endothermic.

Most exothermic: Solution A Middle: Solution B Most endothermic: Solution C The enthalpy of solution depends on the endothermic or exothermic nature of the three processes that produce the overall formation of the solution: separation of solute particles, separation of solvent particles, and formation of solute-solvent interactions. As the forces of attraction become larger, so does the enthalpy component. Based on the given strength of the types of forces, estimate their relative magnitude. In doing so, remember that some processes require energy and others release energy. The dissolution of calcium chloride in water is very exothermic and is used in "hot compresses" to produce heat. When the inner plastic bag of the compress is broken, the calcium chloride dissolves in the water contained in the outer bag. The reaction is very exothermic and produces the desired warming effect.

Draw the Lewis structure for NO.

N double bonded to O. N has 3 valence electrons and O has 4. Although there is no formal positive charge, this molecule is electron deficient because nitrogen does not have a complete octet because of the radical electron.

Consider the following hypothetical molecular collisions and predict which of the following will form potential products, given the values for the energy of activation, Ea, and enthalpy of the reaction, ΔH, in combination with the molecular orientation. Check all that apply. NO2 --> + <---FF Ea = 2.48kJ, ΔH=-5.98 kJ <--ClO + ClO (moving NE) Ea = 0, ΔH = +1.25kJ <--NO + Cl2--> Ea = 4.58 kJ, ΔH = +5.38 kJ NOCl--> + <--NOCl Ea = 6.36 kJ, ΔH = -3.58 kJ NO2 (moving SW) + CO (moving SE) Ea = 7.55 kJ, ΔH = +9.56 kJ NO (SE) + O3 (<--) Ea = 0.59 kJ, ΔH = -1.28 kJ

NO2 --> + <---FF Ea = 2.48kJ, ΔH=-5.98 kJ NOCl--> + <--NOCl Ea = 6.36 kJ, ΔH = -3.58 kJ NO (SE) + O3 (<--) Ea = 0.59 kJ, ΔH = -1.28 kJ Reactions in which Ea>ΔH are either endothermic, when ΔH is positive, or exothermic, when ΔH is negative. Reactants must also be properly orientated for a reaction to occur. When molecules are properly oriented and there is sufficient energy to break bonds products may be formed. First compare the enthalpy of reaction, ΔH, with the energy of activation, Ea, for the reaction and then determine whether the reaction is possible or impossible. Consider that if ΔH is greater than the energy of the transition state, the reaction will not occur. Similarly, Ea must be greater than or equal to zero for a reaction to occur. Also, consider that the reaction can only occur when reactant molecules are properly oriented such that initial bonds are broken and new chemical bonds are formed. For a reaction to be considered possible, the energy of activation, Ea, is greater than the enthalpy of reaction, ΔH.

The addition of NO accelerates the decomposition of N2O, possibly by the following mechanism: NO(g) + N2O(g) → N2(g) + NO2(g) 2NO2(g) → 2NO(g) + O2(g) If experiments show that during the decomposition of N2O, NO2 does not accumulate in measurable quantities, does this rule out the proposed mechanism

No Since NO2 is produced and then consumed during the reaction, it is an intermediate, which explains why NO2 fails to accumulate to a measurable quantity. As a result, NO2 is not present in the balanced chemical equation.

The formation of 2 NO2(g) from NO(g) and O2(g) is believed to occur by a two-step mechanism: NO(g) + O2(g) → NO3(g) (fast) NO3(g) + NO(g) → 2 NO2(g) (slow) Write the rate law for the overall reaction.

Rate = k[NO]^2[O2] We are given a mechanism with a fast initial step and asked to write the rate law for the overall reaction. The rate law of the slow elementary step in a mechanism determines the rate law for the overall reaction. Thus, we first write the rate law based on the molecularity of the slow step. In this case, the slow step involves the intermediate NO3 as a reactant. Experimental rate laws, however, do not contain the concentrations of intermediates; instead they are expressed in terms of the concentrations of reactants and, in some cases, products. Thus, we must relate the concentration of NO3 to the concentration of NO and O2 by assuming that an equilibrium is established in the first step. In doing so, the rate law for the overall reaction is Rate = k[NO][NO][O2] = k[NO]^2[O2].

Consider the following elementary steps that make up the mechanism of a certain reaction: 2X→Y+Z Y+2L→M+Z What is the rate law for step 1 of this reaction?

Rate = k[X]^2 The molecularity of the reaction determines the overall order. The order with respect to each reactant is determined by looking at the number of molecules of each reactant in the equation.

Consider the following elementary steps that make up the mechanism of a certain reaction: 2X→Y+Z Y+2L→M+Z What is the rate law for step 2 of this reaction?

Rate = k[Y][L]^2 The molecularity of the reaction determines the overall order. The order with respect to each reactant is determined by looking at the number of molecules of each reactant in the equation.

Based on their activation energies and energy changes and assuming that all collision factors are the same, rank the following reactions from slowest to fastest. Ea = 33 kJ/mol, ΔE = −12 kJ/mol Ea = 46 kJ/mol, ΔE = −26 kJ/mol Ea = 60 kJ/mol, ΔE = 12 kJ/mol

Slowest reaction: Ea= 60 kJ/mol, ΔE = 12 kJ/mol Medium reaction: Ea= 46 kJ/mol, ΔE = −26 kJ/mol Fastest reaction: Ea= 33 kJ/mol, ΔE = −12 kJ/mol The rate constant determines how fast a reaction proceeds. The magnitude of the activation energy, Ea, influences the rate constant. The lower the value of Ea, the faster the reaction will proceed. A lower activation energy means that, at a given temperature, there will be more molecules with sufficient energy for the reaction to proceed. With a higher activation energy there would be fewer molecules with sufficient energy for the reaction to proceed. The overall change in energy for the reaction, ΔE, has no effect on the reaction rate and therefore has no effect on how fast the reaction proceeds.

What is the molecularity of an elementary step, given its rate law is as follows: Rate = k[A]^2[B]?

Termolecular

https://session.masteringchemistry.com/problemAsset/3236342/1/Figure_P13.6.jpg If the partial pressure of a gas over a solution is tripled, how has the concentration of gas in the solution changed after equilibrium is restored?

The concentration of gas in the solution would triple. The solubility of a gas in a liquid solvent increases in direct proportion to the partial pressure of the gas above the solution.

In a certain hypothetical experiment, the initial concentration of the reactant R is 1.00 mol⋅L−1 , and its rate constant is 0.0150 mol⋅L−1⋅s−1. It follows a zero-order reaction mechanism for the consumption of reactant R. Plot the graph of concentration versus time. Consider the time intervals as 0, 10, 20, 30, 40, and 50 s. Then select the appropriate graph and plot the appropriate points.

The concentration values from t=0 through t=50 are: 1, .85, .7, .55, .4, .25 The graph can be seen here: https://session.masteringchemistry.com/problemAsset/1244710/37/MC_1127418_D.F.jpg The graph of concentration versus time for a typical zero-order reaction is a straight line with a negative slope. The slope of the line is equal to −k and the y intercept is [R]0. There is a steady decrease in concentration until the reaction stops. To plot a graph of concentration versus time, determine the concentration of R at the given time intervals using the integrated rate law. The integrated rate law for the given zero-order reaction is: [R]t=−kt+[R]0 where [R]t is the concentration at time t, k is the reaction constant, t is time, and [R]0 is the initial concentration. For the reaction, [R]0 is 1.00 mol⋅L−1 and k is 0.0150 mol⋅L−1⋅s−1. By knowing the concentration of R at the given time intervals, you can plot a graph of concentration versus time. Make sure to plot the concentration values on the y axis and the corresponding time on the x axis for all of the given time intervals.

Given that dashed lines indicate the weak bonds that are beginning to form, which is the most probable transition state for the formation of NOF? https://session.masteringchemistry.com/problemAsset/1244706/22/BLB-1070911_choice1.jpg https://session.masteringchemistry.com/problemAsset/1244706/22/BLB-1070911_choice2.jpg https://session.masteringchemistry.com/problemAsset/1244706/22/BLB-1070911_choice3.jpg https://session.masteringchemistry.com/problemAsset/1244706/22/BLB-1070911_choice4.jpg

The diagram with the dashed line connecting the F2 molecule to the free radical electron on the N. The most reactive electrons are usually the most unstable. Which electron(s) from NO form the bond with F2? Those in the radical. In this reaction, the radical electron from NO reacts with one of the electrons from the bond between the fluorine atoms in F2, resulting in NOF and a fluorine atom with a radical electron.

Consider two reactions. Reaction (1) has a constant half-life, whereas reaction (2) has a half-life that gets longer as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

The first reaction is a first order reaction, the second one should have an order greater than first.

Your lab partner is flustered. In front of him are two beakers filled with colorless, odorless liquid. They have no labels, except one has a blue piece of tape on it and the other has a pink piece of tape. Your lab partners tells you, "One of these is a solution of pure potassium nitrate in water, and the other one is a colloid of starch in water. But I have no idea which is which. I used up all my reagents, and I have to get to the next step in my reaction. Can you help me figure out which is which?" Being a wise chemist, you immediately figure out an easy way to solve your lab partner's dilemma. You pull out your laser pointer from your pocket and aim the beam through both beakers in order. When you aim at the bottle with the pink label, you see the laser beam going through the liquid very clearly. When you point the laser through the beaker with the blue label, you don't see any laser beam inside the liquid. You see a bright spot on the wall though. Immediately, you announce to your lab partner the following answer: I don't know. The blue labeled bottle contains the starch colloid. The pink labeled bottle contains the starch colloid. The pink labeled bottle contains the potassium nitrate.

The pink labeled bottle contains the starch colloid. The problem deals with how light interacts with different mixtures. If the particles in a mixture are big enough, then light will be scattered by these particles, making the light beam visible. An example of this is when the light show shines through the fog in a rock concert. This fog is a colloidal suspension of particles in the air.

The rate of reaction in terms of the "rate law expression" includes the rate constant (k), the concentration of the reactants, and the orders of the reaction with respect to the different reactants. Consider the following reaction: A+B→C+D The initial concentrations of the reactants A and B are 0.400 M and 0.400 M, respectively. The rate of reaction is 0.060 M⋅s−1, and the orders of the reaction, with respect to reactants A and B, are 1 and 2, respectively. Determine the rate constant (k) for the reaction using the rate law.

The rate constant(k) = 0.938 M^−2⋅s^−1 Using the rate law expression, rate=k[A]^m[B]^n the rate constant k is calculated as k = rate/([A]^m[B]^n) = 0.060 M⋅s^−1 / ((0.400 M)^1×(0.400 M)^2) = 0.938 M−2⋅s−1 First, identify how the rate of reaction can be expressed by using the rate law expression. To do this, you can use the interactive activity. Click on the "Rate law expression" button to see how the rate of reaction is expressed in terms of the rate constant (k), the concentrations of reactants, and the orders of the reaction with respect to the different reactants. Once you know the rate law expression, substitute the values of the rate of reaction, the initial concentrations of reactants A and B, and the order of reactants A and B to calculate the rate constant (k). The rate law expression gives the relationship between the rate of reaction and the concentration of the reactants. For a general reaction, A + B→C + D the rate law expression is rate=k[A]m[B]n where k = the rate constant, and m and n are the orders of the reaction with respect to reactants A and B.

The rate of the reaction in terms of the "appearance of product" includes the change in the concentration of the product, the time interval, and the coefficient of the product. Consider the following reaction: 2A+3B→3C+2D The concentrations of product C at three different time intervals are given. Use the following data to determine the rate of reaction in terms of the appearance of product C between time = 0 s and time = 20 s. Time (s) 0 _______ 20 _______ 40 [C](M) 0.000 ___ 0.0480 ___ 0.0960 Express your answer in molar concentration per second to three significant figures.

The rate of reaction = 8.00×10^−4 M⋅s−1 For the reaction 2A+3B→3C+2D, the rate of the reaction in terms of the disappearance of products C is: (1/3)Δ[C]/t = (1/3)(([C]20 − [C]0)/t) = (1/3)((0.0480 M − 0.0800 M)/20 s) = 8.00×10−4 M⋅s−1 First, identify how the rate of reaction can be expressed in terms of the appearance of product C. To do this, you can use the interactive activity. Click on the "Appearance of products" button to see how the rate of reaction can be expressed in terms of the appearance of product C for the given reaction. Now, determine the change in the concentration of product C (Δ[C]) during the time interval t, where t = 0 to 20 s by using the values of [C] given in the table. If [C]20 is the concentration of product C at t = 20 s and [C]0 is the concentration of the product C at t = 0 s, then the change in concentration of product C is: Δ[C]=[C]20−[C]0 Once you know Δ[C], t, and the expression for the rate of reaction in terms of C, you can calculate the rate of reaction.

The rate of the reaction in terms of the "disappearance of reactant" includes the change in the concentration of the reactant, the time interval, and the coefficient of the reactant. Consider the following reaction: 2A+3B→3C+2D The concentrations of reactant A at three different time intervals are given. Use the following data to determine the average rate of reaction in terms of the disappearance of reactant A between time = 0 s and time = 20 s. Time (s) 0 _________ 20 _______ 40 [A](M) 0.0800 ___ 0.0480 ___ 0.0360 Express your answer in molar concentration per second to three significant figures.

The rate of reaction = 8.00×10^−4 M⋅s−1 For the reaction 2A+3B→3C+2D, the rate of the reaction in terms of the disappearance of reactants A is: −(1/2)Δ[A]/t = −(1/2)(([A]20 − [A]0)/t) = −(1/2)((0.0480 M − 0.0800 M)/20 s) = 8.00×10−4 M⋅s−1 First, identify how the rate of reaction can be expressed in terms of the disappearance of reactant A. To do this, you can use the interactive activity. Click on the "Disappearance of reactants" button to see how the rate of reaction can be expressed in terms of the disappearance of reactant A for the given reaction. Now, determine the change in the concentration of reactant A (Δ[A]) during the given time interval t where t = 0 to 20 s by using the values of [A] given in the table. If [A]20 is the concentration of reactant A at t = 20 s, and [A]0 is the concentration of the reactant A at t = 0 s, then the change in concentration of reactant A is: Δ[A]=[A]20−[A]0 Once you know Δ[A], t, and the expression for the rate of reaction in terms of A, you can calculate the rate of reaction.

Consider the following reaction: O2+2NO→2NO2, rate = k[O2][NO]^2 What would happen to the rate if [O2] were doubled?

The rate would double. The reaction order with respect to O2 is First-Order. For the equation y = x^1, what happens to the value of y when x is doubled? For a particular reaction, aA+bB+cC→dD, the rate law was experimentally determined to be rate = k[A]^0[B]^1[C]^2 =k[B][C]^2 This equation is zero order with respect to A. Therefore, changing the concentration of A has no effect on the rate because [A]^0 will always equal 1. This equation is first order with respect to B. This means that if the concentration of B is doubled, the rate will double. If [B] is reduced by half, the rate will be halved. If [B] is tripled, the rate will triple, and so on. This equation is second order with respect to C. This means that if the concentration of C is doubled, the rate will quadruple. If [C] is tripled, the rate will increase by a factor of 9, and so on.

Consider the following reaction: O2+2NO→2NO2, rate = k[O2][NO]^2 What would happen to the rate if [NO] were doubled?

The rate would quadruple. The reaction order with respect to O2 is First-Order. For the equation y = x^1, what happens to the value of y when x is doubled? For a particular reaction, aA+bB+cC→dD, the rate law was experimentally determined to be rate = k[A]^0[B]^1[C]^2 =k[B][C]^2 This equation is zero order with respect to A. Therefore, changing the concentration of A has no effect on the rate because [A]^0 will always equal 1. This equation is first order with respect to B. This means that if the concentration of B is doubled, the rate will double. If [B] is reduced by half, the rate will be halved. If [B] is tripled, the rate will triple, and so on. This equation is second order with respect to C. This means that if the concentration of C is doubled, the rate will quadruple. If [C] is tripled, the rate will increase by a factor of 9, and so on.

In a study of the decomposition of the compound X via the reaction X(g)⇌Y(g)+Z(g) the following concentration-time data were collected: Time (min) [X](M) 0 _______ 0.467 1 _______ 0.267 2 _______ 0.187 3 _______ 0.144 4 _______ 0.117 5 _______ 0.099 6 _______ 0.085 7 _______ 0.075 What is the order of the reaction?

The reaction is *second* order. From the concentration-time plots we can determine the order of the reaction and also the rate constant (which is represented by the slope of the straight line). For a zeroth-order reaction, the plot of [A] versus time is a straight line. For a first-order reaction, the plot of ln[A] versus time is a straight line. If a reaction is second order, the plot of 1/[A] versus time is a straight line. Draw the plots and see which one is linear. It will tell you what the reaction order is.

When ammonium chloride dissolves in water, the solution becomes colder. Why does the solution form? The solution forms because the favorable entropy of mixing outweighs the *increase* in enthalpy by the solution. The solution forms because the favorable entropy of mixing outweighs the *decrease* in enthalpy by the solution.

The solution forms because the favorable entropy of mixing outweighs the *increase* in enthalpy by the solution. Since the reaction is spontaneous, it can be implied that the favorable entropy effects outweigh the increase in enthalpy, forming the solution.

Indicate whether each statement is true or false. When a solution is made, the *entropy* of mixing is always a *negative* number. A solute will dissolve in a solvent if solute-solute interactions are *weaker* than solute-solvent interactions. A *decrease* in entropy favors mixing.

True: A solute will dissolve in a solvent if solute-solute interactions are weaker than solute-solvent interactions. False: When a solution is made, the entropy of mixing is always a negative number. A decrease in entropy favors mixing. For a solution to mix the solute-solute interactions as well as the solvent-solvent interactions must be broken. This requires energy and, therefore, is considered to be an endothermic process, which means the enthalpies for these steps will be positive. The solute-solvent interactions then form and release energy, making this step exothermic with a negative enthalpy. The enthalpy for the formation of the solution is the sum of the three enthalpies ΔHsoln=ΔHsolute+ΔHsolvent+ΔHmix and can be positive or negative depending on the relative magnitude of the individual enthalpies for the steps. Solution formation is favored by an increase in entropy and a decrease in enthalpy (negative enthalpy change).

Indicate whether each statement is true or false. NaCl dissolves in water but not in hexane because the enthalpy of mixing NaCl with water is more negative than that of mixing NaCl with hexane. NaCl dissolves in water but not in benzene (C6H6) because benzene is denser than water. NaCl dissolves in water but not in benzene because the water-ion interactions are stronger than benzene-ion interactions.

True: NaCl dissolves in water but not in benzene because the water-ion interactions are stronger than benzene-ion interactions. NaCl dissolves in water but not in hexane because the enthalpy of mixing NaCl with water is more negative than that of mixing NaCl with hexane. False: NaCl dissolves in water but not in benzene (C6H6) because benzene is denser than water. Nonpolar substances are more likely to dissolve in nonpolar solvents. Ionic and polar substances are more likely to dissolve in polar solvents. This is due to the fact that ions and polar substances will have stronger intermolecular interactions with each other since they can have ion-dipole or dipole-dipole attractions, while nonpolar substances only interact with weaker dispersion forces. This is often generalized as "like dissolves like." NaCl will dissolve in polar solvents, water or ethanol, and will not dissolve in nonpolar solvents, hexane or benzene, because of the ion-dipole attractions between Na+ or Cl− and polar substances. The density of the solvent does not play an important role in determining if substances will mix.

You make a solution of a nonvolatile solute with a liquid solvent. Indicate whether each of the following statements is true or false. The *boiling point* of the solution is *lower* than that of the pure solvent. The *freezing point* of the solution is *higher* than that of the pure solvent. The *freezing point* of the solution is *lower* than that of the pure solvent. The *boiling point* of the solution is *higher* than that of the pure solvent.

True: The *boiling point* of the solution is *higher* than that of the pure solvent. The *freezing point* of the solution is *lower* than that of the pure solvent. False: The *boiling point* of the solution is *lower* than that of the pure solvent. The *freezing point* of the solution is *higher* than that of the pure solvent.

On a reaction energy profile diagram, how is an intermediate represented? peak downward peak valley upward peak

Valley

A zero-order reaction has a constant rate of 3.00×10^−4 M/s. If after 50.0 seconds the concentration has dropped to 2.50×10^−2 M, what was the initial concentration?

[A]0 = 4.00×10^−2 M First, choose the appropriate integrated rate law equation based on the reaction order. Then plug the given values into the formula and solve for [A]0. For a zero-order reaction, rate=k[A]0.

A zero-order reaction has a constant rate of 4.50×10^−4 M/s. If after 45.0 seconds the concentration has dropped to 5.00×10^−2 M, what was the initial concentration?

[A]0 = 7.03×10^−2 M First, choose the appropriate integrated rate law equation based on the reaction order. Then plug the given values into the formula and solve for [A]0. For a zero-order reaction, rate = k[A]0 = k. Therefore for this reaction the rate constant is 4.50^×10^−4 M/s. The order is zero-order therefore to solve for [A]0 use: [A] = [A]0 −kt

The rate constant for a certain reaction is k = 8.50×10^−3 s−1 . If the initial reactant concentration was 0.900 M, what will the concentration be after 5.00 minutes?

[A]t = 7.03×10^−2 M First, determine the reaction order because that will indicate which integrated rate law to use. To determine the reaction order, notice that the units of the rate constant for this reaction are s−1. The units of the rate constant can be used to determine the overall order of the reaction based on the rate law. Consider the following rate law. rate = k[A]^x M/s = s−1 × M^x Identify what x must be for the units to come out correctly. Once you have the appropriate rate law, plug the given values into that formula and solve for [A]. Don't forget to convert minutes to seconds (to match the units of k).

The rate constant for a certain reaction is k = 3.80×10^−3 s−1 . If the initial reactant concentration was 0.950 M, what will the concentration be after 20.0 minutes?

[A]t = 9.94×10^−3 M First, determine the reaction order because that will indicate which integrated rate law to use. To determine the reaction order, notice that the units of the rate constant for this reaction are s−1. The units of the rate constant can be used to determine the overall order of the reaction based on the rate law. Consider the following rate law. rate = k[A]^x M/s = s−1×M^x Identify what x must be for the units to come out correctly. Once you have the appropriate rate law, plug the given values into that formula and solve for [A]. Don't forget to convert minutes to seconds (to match the units of k). The order of this reaction is First-Order. Therefore use either of the following formulas to solve for [A]t: [A]t = [A]0*e^−kt or ln[A]t - ln[A]0 = -kt

[X] = 4.36×10−2 M Write the integrated rate law for the reaction. The initial concentration, [X]0, is the concentration of X at the very beginning of the reaction, when t=0 (which you can find in the data table). Solve this equation for [X].

https://session.masteringchemistry.com/problemAsset/1244768/11/uncatalyzed.jpg What is the value of the activation energy of the uncatalyzed reaction?

activation energy = 175 kJ The activation energy of a reaction is the difference in energy between the reactants and the activated complex. Activation energy is always a positive number.

https://session.masteringchemistry.com/problemAsset/1244768/11/uncatalyzed.jpg What is the value of the activation energy of the uncatalyzed reaction in reverse?

activation energy = 300 kJ The activation energy of a reaction is the difference in energy between the reactants and the activated complex. However, for the reverse reaction, the reactants and products are switched.

Which of the following would lead to an increased rate of reaction for gaseous reactants? increasing the volume of the reaction container increasing the temperature of the reactants decreasing the partial pressures of the reactants decreasing the concentration of the reactants

increasing the temperature of the reactants On a molecular level, reaction rates depend on the frequency of collisions. The greater the frequency of collisions, the higher the reaction rate.

Consider the data presented below. Time (s): ____ 0 _____ 40 _____ 80 ____ 120 ___ 160 Moles of A: _ 0.100 __ 0.067 __ 0.045 __ 0.030 __ 0.020 What is the rate constant?

k = 1.01×10^−2 s−1 (Use ln([A]t/[A]0) = -kt to find the value of k. Make sure to use the first and last data point to get as accurate of an answer as possible.)

The gas-phase reaction of NO with F2 to form NOF and F has an activation energy of Ea = 6.30 kJ/mol and a frequency factor of A = 6.00×10^8 M−1⋅s−1 . What is the rate constant at 513 ∘C ?

k = 2.29×10^8 M−1⋅s−1 Use either lnk = (-Ea/RT) + lnA or The Arrhenius equation relates the rate constant, k, to the Kelvin temperature, T: k=Ae^(−Ea/RT) where A is the frequency factor, Ea is the activation energy in joules, and R = 8.314 J/(mol⋅K) is the gas constant. The lowercase e symbol refers to the number 2.718, the base of the natural logarithm in mathematics. Most scientific calculators have an e^x button available, but typing in the value also works.

k = 3.0×10^−2 M−2⋅s−1 Based on your previous answers, rate=k[A]^2[B]^0[C]^1. Now, plug in numbers from any one of the trials and solve for k. To determine the units for k, use dimensional analysis. Cross out like units in the numerator and denominator of the expression until you reduce the unit to terms that cannot be canceled out. To check your units, set up the rate equation using the units determined for k, and the units of concentration for each reagent. Use dimensional analysis to verify that the rate will be calculated in M/s.

The reactant concentration in a zero-order reaction was 0.100 M after 135 s and 2.00×10^−2 M after 325 s . What is the rate constant for this reaction? What was the initial reactant concentration for the reaction described in Part A?

k0th = 4.21×10^−4 M/s [A]0 = 0.157 M The graph of concentration versus time yields a straight line with a slope of −k (for zero-order reaction). Therefore if you treat the given values as point coordinates, you can calculate the slope of the line formed by connecting those points. i.e. rise is the difference in concentration and run is the difference in time. Alternatively you can just plug the values in the integrated rate law for a zero-order reaction, [A] = −kt + [A]0, and solve for k. Use the integrated rate law for zero-order reactions [A] = −kt + [A]0 to solve for [A]0. Values for [A] and t are given in Part A, and your answer to Part A was the value of k. You can use either data point given in Part A to complete the calculation, however the corresponding time t and concentration [A] must be used together.

The reactant concentration in a first-order reaction was 6.20×10^−2 M after 10.0 s and 8.50×10^−3 M after 90.0 s . What is the rate constant for this reaction?

k1st = 2.48×10^−2 s−1 (Use the integrated rate law for a first-order reaction: ln[A]=−kt+ln[A]0)

The reactant concentration in a second-order reaction was 0.740 M after 290 s and 6.20×10^−2 M after 710 s . What is the rate constant for this reaction?

k2nd = 3.52×10^−2 M−1⋅s−1 (Use the integrated rate law for a second-order reaction: 1/[A]= kt+1/[A]0. Pay attention to the units.)

Consider the following multistep reaction: C+D⇌CD(fast) CD+D→CD2(slow) CD2+D→CD3(fast) ---------------------- C+3D→CD3(overall) Based on this mechanism, determine the rate law for the overall reaction. Express your answer in standard MasteringChemistry format. For example, if the rate law is k[C]3[D]2 type k*[C]^3*[D]^2.

k[C][D]^2 A multistep reaction can only occur as fast as its slowest step. Therefore, it is the rate law of the slow step that determines the rate law for the overall reaction. Because CD is an intermediate [CD] must be replaced with what it equals in the first step [C][D]

A first-order reaction will yield a straight line when time is plotted on the x-axis and _______ is plotted on the y-axis.

ln[A]t

An average reaction rate is calculated as the change in the concentration of reactants or products over a period of time in the course of the reaction. An instantaneous reaction rate is the rate at a particular moment in the reaction and is usually determined graphically. The reaction of compound A forming compound B was studied and the following data were collected: Time (s) ____ [A] (M) 0. _________ 0.184 200. _______ 0.129 500. _______ 0.069 800. _______ 0.031 1200. ______ 0.019 1500. ______ 0.016 What is the average reaction rate between 0. and 1500. s?

rate = 1.12×10^−4 M/s Rate = (change in concentration) / (change in time)

rate = 6.8×10^−5 M/s You'll need to plot the points on a graph, draw a line tangent to the point at 800s, and find the slope. Slope = Δy/Δx or you can draw the tangent line to the y and x axes in which case the slope will be (point where tangent line intersects y-axis) / (point where tangent line intersects x-axis).

rate = 8.69×10^−5 M/s

Consider the reaction A + 2B ⇌ C whose rate at 25 ∘C was measured using three different sets of initial concentrations as listed in the following table: Trial ___ [A](M) ___ [B](M) ___ Rate(M/s) 1 ______ 0.40 ____ 0.050 ____ 9.6×10^−3 2 ______ 0.40 ____ 0.100 ____ 1.9×10^−2 3 ______ 0.80 ____ 0.050 ____ 3.8×10^−2 What is the rate law for this reaction? Express the rate law symbolically in terms of k, [A], and [B].

rate = k[A]^2[B] The rate law is an equation that describes the dependence of the reaction rate on the concentration of each reactant. For the given reaction, the general expression would be rate=k[A]m[B]n The exponents m and n can be determined by comparing the trails and finding how the change in the reactant concentrations affected the reaction rate. m is the reaction order for A, and n is the reaction order for B. To determine the reaction order for each, consider the following definitions: A zeroth-order relationship means that the rate does not depend on concentration That is, for any [A] value you would see [A]^0=1 or for any B value you would see [B]^0=1 if this reaction is zeroth order in either. You would see a constant rate even as one or the other changes. A first-order relationship means that the rate is directly proportional to concentration: rate∝[A] or rate∝[B]. For example, if this reaction is first order in [A], you will see the rate double if [A] doubles, or the rate triple if [A] triples, etc. A second-order relationship means that the rate is directly proportional to the square of the concentration: rate∝[A]2 or rate∝[B]2. For example, if this reaction is second order in [A], you will see the rate quadruple if [A] doubles, or the rate increase by a factor of nine if [A] triples. By comparing trials 1 and 3, the reaction order with respect to A can be determined to be second-order. By comparing trials 1 and 2, the reaction order with respect to B can be determined to be first-order.

Metals often form several cations with different charges. Cerium, for example, forms Ce3+ and Ce4+ ions, and thallium forms Tl+ and Tl3+ ions. Cerium and thallium ions react as follows: 2Ce^4+(aq) + Tl+(aq) → 2Ce^3+(aq) + Tl^3+(aq) This reaction is very slow and is thought to occur in a single elementary step. The reaction is catalyzed by the addition of Mn2+(aq) according to the following mechanism: Ce^4+(aq) + Mn^2+(aq) → Ce^3+(aq) + Mn^3+(aq) Ce^4+(aq) + Mn^3+(aq) → Ce^3+(aq) + Mn^4+(aq) Mn^4+(aq) + Tl+(aq) → Mn^2+(aq) + Tl^3+(aq) What is the rate law for the uncatalyzed reaction?

rate = k[Ce4+]^2[Tl+] This means that the rate of the reaction is dependent on two cerium ions and one thallium ion all colliding with the correct energy and orientation at the same time. For a reaction that undergoes a single elementary step, the coefficients of the reactants determine the rate law. In these cases, the coefficients are the number of each atom that must collide at the same time with the right angle and energy for the reaction to occur. This means that the reaction order of each reactant, and thus the exponent in the rate law, is equal to the coefficient of that reactant in the equation. For example, for an elementary step of the form aA+bB→cC+dD where the lowercase letters are the coefficients and the uppercase letters are the molecular formulas, the rate law is rate=k[A]^a[B]^b.

Consider the reaction of hydrogen peroxide with iodine: H2O2(aq) + I2(aq) ⇌ OH−(aq) + HIO(aq) The reaction is first order in I2 and second order overall. What is the rate law?

rate = k[H2O2][I2] The reaction order exponents indicate how sensitive the rate is to changes in the concentrations of reactants. For simple reactions, the exponents are usually small positive integers. For more complex reaction, the exponents can be negative, zero, or even fractional. When the order is zero, the rate is independent of the concentration of the reactant. When the order is negative, the rate decreases as the concentration increases. For this case, first order in H2O2 means that the reaction rate depends linearly on the concentration of H2O2. For example, if [H2O2] is doubled, the rate doubles (increases by a factor of 2). The same is true for I2.

You study the rate of a reaction, measuring both the concentration of the reactant and the concentration of the product as a function of time, and obtain the following results: https://session.masteringchemistry.com/problemAsset/3237508/1/MC_2537913_graph.jpg Choose equivalent expressions for the rate of the reaction in terms of the appearance or disappearance of the two substances. rate = −(1/2)Δ[B]/Δt = Δ[A]/Δt rate = −Δ[B]/Δt = (1/2)Δ[A]/Δt rate = −Δ[B]/Δt = Δ[A]/Δt rate = Δ[B]/Δt = Δ[A]/Δt rate = (1/2)Δ[B]/Δt = Δ[A]/Δt

rate = −Δ[B]/Δt = (1/2)Δ[A]/Δt For the reaction B→2A the rate is given by the following rate law: rate = −Δ[B]/Δt = (1/2)Δ[A]/Δt Because [B] decreases, Δ[B] is a negative number. However, by convention, rates are always expressed as positive quantities. Thus, the minus sign in the equation converts the negative Δ[B] to a positive rate of disappearance. Because 2 mol of A form for each mole of B that reacts, the rate of appearance of A is twice the rate of disappearance of B. Therefore, divide the rate of appearance of A by 2 (its coefficient in the balanced chemical equation).

In the hydrogenation of ethylene using a nickel catalyst, the initial concentration of ethylene is 1.50 mol⋅L−1 and its rate constant (k) is 0.0018 mol⋅L−1⋅s−1 . Determine the rate of reaction if it follows a zero-order reaction mechanism.

rate of reaction = 1.8×10^−3 mol⋅L−1⋅s−1 (<-- units are a long way of saying M/s) To determine the rate of the given reaction, consider that the reaction follows a zero-order reaction mechanism. For zero-order reactions, the rate of reaction is independent of the concentration of the reactants. Based on this, identify the rate law for a zero-order reaction. By knowing the rate law, the rate of the reaction can be determined. The rate law for a zero-order reaction is rate = k[A]0 = k

How does the magnitude of ΔHmix compare with the magnitude of ΔHsolvent+ΔHsolute for endothermic solution processes? The magnitude of ΔHmix for an endothermic solution will be __________ than the magnitude of _______________ https://session.masteringchemistry.com/problemAsset/3236340/1/fig2.jpg

smaller ΔHsolvent+ΔHsolute

In a saturated solution of a salt in water seed crystal addition may cause massive crystallization the rate of crystallization = the rate of dissolution addition of more water causes massive crystallization the rate of dissolution > the rate of crystallization the rate of crystallization > the rate of dissolution

the rate of crystallization = the rate of dissolution

https://media.pearsoncmg.com/bc/bc_0media_chem/legacy_media/int_act_ener_solu/index.html Select the steps that are associated with energy entering the system. Check all that apply. the separation of solvent molecules the separation of solute molecules the mixing of solute and solvent molecules

the separation of solvent molecules the separation of solute molecules In the diagram, enthalpy and energy are increasing moving from the bottom to the top of the plot. Upon clicking the diagram, you can either read the descriptions outlined in yellow to determine which steps involve energy entering the system or you can look at the direction of the red arrow in each step. Steps that involve an increase in energy are indicated by an arrow pointing upward. Steps associated with energy entering the system are endothermic and have a a positive enthalpy value. Steps that release energy into the surroundings are exothermic and have a negative enthalpy value.

https://session.masteringchemistry.com/problemAsset/1244768/11/uncatalyzed.jpg What is the value of the enthalpy change of the uncatalyzed reaction?

ΔH = -125 kJ (Note the negative sign) The enthalpy change of a reaction is the difference in energy between the reactants and the products: ΔH = Eproducts − Ereactants

https://session.masteringchemistry.com/problemAsset/1244768/11/uncatalyzed.jpg What is the value of the enthalpy change of the uncatalyzed reaction in reverse?

ΔH = 125 kJ The enthalpy change of a reaction is the difference in energy between the reactants and the products: ΔH=Eproducts−Ereactants However, for the reverse reaction, the reactants and products are switched.

https://media.pearsoncmg.com/bc/bc_0media_chem/legacy_media/int_act_ener_solu/index.html A student working on a laboratory experiment dissolves a hypothetical solute AB in water. The student observes that the flask becomes cold to the touch. Use the diagram to determine which of the following equations accurately shows the relationship between the magnitude of the enthalpy values for each step in the dissolution of AB in water for a flask that becomes cold to the touch. ΔH1 + ΔH2 > ΔH3 ΔH1 + ΔH2 = ΔH3 ΔH1 + ΔH2 < ΔH3

ΔH1 + ΔH2 > ΔH3 First determine the direction in which energy is flowing in the system. Then determine what values for ΔH1, ΔH2, and ΔH3 will result in that energy flow. Recall from Part A that both ΔH1 and ΔH2 are associated with an absorption of energy and ΔH3 is associated with a release of energy. When solute AB dissolves in water the flask becomes cold to the touch. What direction is energy flowing in the system as solute AB dissolves in water? Energy is flowing into the system from the surroundings. The enthalpy value for this reaction is positive. Using the diagram determine which situation results in a positive ΔHsoln. ΔHsoln is positive when the solute-solute and solvent-solvent interactions are greater than the solute-solvent interactions. When |ΔH1+ΔH2|>|ΔH3| the value of ΔHsoln is positive. The reaction is endothermic.

https://media.pearsoncmg.com/bc/bc_0media_chem/legacy_media/int_act_ener_solu/index.html Select the ΔH values associated with the dissolution of lithium chloride that are exothermic. Check all that apply. ΔH1: energy associated with the separation of water molecules ΔH2: energy associated with the separation of LiCl ions ΔH3: energy associated with the formation of water-ion interactions ΔHsoln: the enthalpy of solution

ΔH3: energy associated with the formation of water-ion interactions ΔHsoln: the enthalpy of solution Exothermic processes are associated with a negative ΔH value.

Ch 13 & 14

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